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AP Physics Chapter 4 Full Notes
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AP Physics - Chapter 4

Lesson 16: Forces & Free-Body Diagrams

Notes

LEARNING TARGET & JOURNALS

  • I can construct and interpret free-body diagrams.

NOTES

  • Force is defined as a cause of acceleration.
  • So a change in acceleration can happen in one of two ways:
  1. Change in speed
  2. Change in direction
  • Since acceleration is a vector quantity, so also is force.
  • The variable for force is F.
  • The SI unit for force is the Newton.
  • Which is equivalent to a {"code":"$$kg\\cdot\\frac{m}{s^{2}}$$","font":{"color":"#000000","family":"Calibri","size":11},"aid":null,"id":"3","backgroundColorModified":false,"type":"$$","backgroundColor":"#ffffff","ts":1634826357611,"cs":"MQNj4CORG5WtBaF+06F4uA==","size":{"width":41,"height":26}}

  • A contact force is a force that arises from the physical contact of two objects.
  • A field force is a force that can exist between objects in the absence of direct physical contact.

  • A free body diagram is a diagram of an object showing all forces acting on that object isolated from the rest of the objects in the system.
  1. All forces originate from the center of mass of the object.
  2. The vector to represent the force of gravity (Fg) always points straight down.
  3. The vector to represent the force of friction (Ff) is parallel to the motion of the object, but opposite in direction.
  4. When an object is in contact with a surface, there is a normal force (FN) that is directed perpendicular to the surface.

                

EXAMPLE 1 – Draw a Free-Body Diagram for the following.

  1. A book held in your hand.

  1. A book pushed across the desk

  1. A ball that was just dropped

  1. A car in neutral coasting down a hill

AP Physics - Chapter 4

Lesson 17: Newton’s 1st Law of Motion and Equilibrium

Notes

LEARNING TARGET & JOURNALS

  • I can describe systems experiencing a balanced or unbalanced net force.

NOTES

Newton’s 1st Law of Motion (Law of Inertia)

  • That is constant speed in a straight line
  • This is also called the Law of Inertia.
  • Inertia is the tendency of an object to resist acceleration.

 

  • Equilibrium is defined as the state in which there is no change in the body’s motion.
  • Equilibrium is achieved when the net external force on an object is equal to 0.
  • That means that the acceleration of the object is 0.
  • Which means
  1. the magnitude of the velocity of the object is constant, and
  2. the direction of the velocity is constant.https://lh4.googleusercontent.com/xY3BOYQ6id1Cm9wzdvoIBCpKPPUS3ugWXgKNEC-VS_4EBBp9-Vz_6WIwiHInxK-kpLMzSivPMXPSuIWPGIWxIhLDO-PN3xHhcT8gpsP4_Qj56P8jjNFKWra99pTspTT2XkPfbtjd

  • Static equilibrium describes an object that is not moving while experiencing a net force equal to zero.
  • Dynamic equilibrium describes an object that is moving at constant velocity while experiencing a net force equal to zero.

EXAMPLE 1 – Determine the net force acting on the object. HINT: Include both magnitude and direction for net force.

  1. An 8.0-N weight has one horizontal rope exerting a force of 6.0 N on it.

EXAMPLE 2 – Determine the magnitude and direction of the necessary force to bring the object to equilibrium.

  1. An object in equilibrium has three forces exerted on it. A 33-N force acts at 90° from the x-axis and a 44-N force acts at 60°. What is the magnitude and direction of the third force?
  1. Five forces act on an object:
  1. 60.0 N at 90.0°
  2. 40.0 N at 0.0°
  3. 80.0 N at 270.0°
  4. 40.0 N at 180.0°, and
  5. 50.0 N at 60.0°.

What is the magnitude and direction of a sixth force that would produce equilibrium?

AP Physics - Chapter 4

Lesson 18: Newton’s 2nd Law of Motion

Notes

LEARNING TARGET & JOURNALS

  • I can describe an object’s acceleration in terms of its mass and the net force acting on it.

NOTES

Newton’s 2nd Law of Motion

  • The acceleration of an object is directly proportional to the net external force acting on the object and inversely proportional to the object’s mass.
  • We will take that mess and shrink it down to one simple formula:          {"code":"$$\\sum_{}^{}F\\,=\\,ma$$","id":"1","backgroundColorModified":false,"backgroundColor":"#ffffff","aid":null,"font":{"color":"#000000","size":11,"family":"Calibri"},"type":"$$","ts":1634825885574,"cs":"fcmkC3pQNJLjymjwopAbGw==","size":{"width":78,"height":20}}
  • Σ means “net” or total.

          http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/fmaill.gifhttp://www.physics4kids.com/files/art/motion_laws2_240x180.gif

  • Objects that are either at rest or moving with constant velocity are said to be in equilibrium
  • Newton’s 1st Law states that for equilibrium: the net external force acting on an object must be equal to zero.

  • When using Newton’s 2nd Law of Motion, the acceleration of the object is calculated with the net force acting on the object.
  • So find the net force first, and then plug into .
  • Force is a vector quantity, and the direction of the net force is also the direction of the acceleration.

Connection to Kinematics

  • Kinematic equations can find the net acceleration needed to find the net force.
  • Or, the net force can find the net acceleration to be used in the kinematic equations to find:
  1. the displacement of the object due to the net force
  2. the final velocity of the object after applying the net force
  3. the time the object travels while the net force is applied

EXAMPLE 1 – Two people are pushing a stalled car. The mass of the car is 1850 kg. One person applies a force of 275 N to the car, while the other applies a force of 395 N acting in the same direction.  A third force of 560 N also acts on the car, but in a direction opposite to that in which the people are pushing.  This force arises because of friction and the extent to which the pavement opposes the motion of the tires.

Find the acceleration of the car.

EXAMPLE 2 –         A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in
15 seconds.

  1. What is the expected acceleration of the car from the 10,000 N force?
  1. What is the actual acceleration of the car from the observed data of Δx and Δt?

EXAMPLE 3 -         When an F-14 airplane takes-off from an aircraft carrier it is literally catapulted off the flight deck. The plane's final speed at take-off is 68.2 m/s. The F-14 starts from rest. The plane accelerates in 2 seconds and has a mass of 29,545 kg. What is the total force that gets the F-14 in the air?

AP Physics - Chapter 4

Lesson 19: Newton’s 3rd Law of Motion, Weight & The Normal Force

Notes

LEARNING TARGET & JOURNALS

  • I can identify action-reaction pairs.

NOTES

Newton’s 3rd Law of Motion

  • We will shrink that tongue twister down to for every action, there is an equal and opposite reaction.
  • Example: You are pushing down on your chair and the chair is pushing up on you. 
  • Weight is a measure of the force of gravity acting on a mass.
  • It is the product of the acceleration of gravity and an object’s mass.                {"aid":null,"id":"2","type":"$$","font":{"size":11,"color":"#000000","family":"Calibri"},"backgroundColor":"#ffffff","code":"$$F_{g}=mg$$","backgroundColorModified":false,"ts":1634825963845,"cs":"OOtxciBpbmLK4MV1U0UaCA==","size":{"width":54,"height":13}}
  • The variable for weight is either  or .
  • Typically the magnitude of the weight is all we need, so weight is positive.
  • Given the weight, the mass can be found by dividing out the acceleration of gravity.                {"backgroundColor":"#ffffff","font":{"family":"Calibri","color":"#000000","size":11},"backgroundColorModified":false,"id":"4","code":"$$m=\\frac{F_{g}}{g}$$","type":"$$","aid":null,"ts":1634826395788,"cs":"WvXqcUfRYrYptIOp6g7rsA==","size":{"width":52,"height":33}}
  • The normal force is the magnitude of force that a flat surface must provide on an object.Image result for normal force example
  • Normal is a mathematical term that means “to be perpendicular to.”
  • The magnitude is best calculated using the following equation for flat surfaces and solving for FN.

{"code":"$$\\sum_{}^{}F_{y}=\\,F_{N}\\,+\\,\\sum_{}^{}F_{up}\\,-\\,mg\\,-\\,\\sum_{}^{}F_{down}=\\,0$$","backgroundColorModified":false,"id":"5","backgroundColor":"#ffffff","type":"$$","font":{"color":"#000000","size":11,"family":"Calibri"},"aid":null,"ts":1634828907408,"cs":"PDjdREng1Iao6kfCPyaGfg==","size":{"width":308,"height":20}}

EXAMPLE 1 – Solve the following as related to weight and other gravitational forces.

  1. A locomotive’s mass is 18,181.81 kg. What is its weight?
  1. A small car weighs 10,168.25 N. What is its mass?
  1. The Space Shuttle travels from launch to 529.2 m in 6.0 seconds. What is the acceleration of the shuttle and how many gʼs is this?

EXAMPLE 2 – Calculate the normal force in the following.

Image result for calculate normal force

  1. Image result for calculate normal force
  1. Image result for calculate normal force

AP Physics - Chapter 4

Lesson 20: Apparent Weight

Notes

LEARNING TARGET & JOURNALS

  • I can apply Newton’s Laws of Motion to accelerated systems.

NOTES

  • The apparent weight of an object is observed when that object is accelerating relative to Earth’s gravitational field.
  • To find the apparent weight, solve for FN by using a free-body diagram and Newton’s 2nd Law of Motion.
  • Because FN is actually the force that your body feels, or what a scale would read. Image result for elevator physics free body diagram

There are 4 scenarios that are often played out in an elevator to describe apparent weight.

  • It may be helpful to draw an independent vector in your FBD to represent the
    direction of acceleration.
  • If the acceleration is upward, substitute a positive value for a.
  • If the acceleration is downward, substitute a negative value for a.
  1. Elevator is at constant speed or at rest.
  • Constant speed indicates that , solve for FN.
  • FN equals the weight.

⇒

  1. Elevator accelerating downward
  • Now ,
    solve for     FN.
  • FN < Fg
    So, we feel lighter!

 ⇒

  1. Elevator accelerating upward
  • Now ,
    solve for     FN.
  • FN > Fg
    So, we feel heavier!

 ⇒

  1. Elevator in free-fall (cable broke!)
  • Now , solve for FN?
  • FN = 0
    So, we feel weightless!

 ⇒

  • The same logic above can be used to solve for the tension in the supporting cable, by replacing FN with T in each equation, and then solve for T.

EXAMPLE 1 –         An 85.0 kg man is standing on a scale in an elevator. The scale reads the normal force in newtons. For each of the following situations, draw a free body diagram that is qualitatively accurate and use
Newton’s 2nd Law to determine what the scale reads (HINT: His apparent weight!)

  1. The elevator is at rest.
  1. The elevator accelerates upward at 2.0 m/s2.
  1. The elevator moves upward with a constant speed of 3.0 m/s.
  1. As it gets to the upper floor, the upwardly moving elevator slows down with an acceleration that has a magnitude of 1.50 m/s2. (Think about what the direction of the acceleration is.)
  1. The elevator then starts to go down with an acceleration of 2.5 m/s2.
  1. As the elevator approaches the bottom floor, it slows down while going down with an acceleration that has a magnitude of 1.8 m/s2.
  1. Suddenly, the cable holding the elevator snaps and the elevator goes into free fall.

AP Physics - Chapter 4

Lesson 21: Incline Planes

Notes

LEARNING TARGET & JOURNALS

  • I can account for the normal force on horizontal and inclined surfaces.

NOTES

  • Objects on an incline require us to tilt the coordinate system so that the incline is now the x-axis.
  • The weight vector will still point straight down on the page.
  • The normal force is now a component of the weight.         {"aid":null,"type":"$$","code":"$$F_{N}=\\,F_{g}\\cos\\theta$$","backgroundColor":"#ffffff","backgroundColorModified":false,"font":{"color":"#000000","family":"Calibri","size":"12"},"id":"6-0","ts":1634829082745,"cs":"51L9yLKAUAXZTupI1C4rRQ==","size":{"width":96,"height":16}}
  • Where 𝜃 is the angle between the incline and the horizontal
  • Which is also the angle between the weight and the normal force.
  • The other component of the weight is the force acting down the incline making the object slide

{"type":"$$","backgroundColor":"#ffffff","code":"$$F_{g,parallel}=\\,F_{g}\\sin\\theta$$","id":"6-1","backgroundColorModified":false,"font":{"size":"12","family":"Calibri","color":"#000000"},"aid":null,"ts":1634829358248,"cs":"3xjjGvaC4PtbilvB6GnEVA==","size":{"width":132,"height":16}}

  • For all incline situations, it is often easiest to establish the positive direction for your frame of reference as the way the object is intended to slide along the incline.
  • So, if you feel the object will slide down the incline, treat down as positive.
  • If your final answer comes out negative, you guessed wrong on which way the object will move on the incline!

3 Common Applications For Inclined Planes

  1. Stationary Block on Incline
  • The sum of all forces acting parallel to the incline’s surface is zero.

{"backgroundColor":"#ffffff","code":"$$\\sum_{}^{}F_{parallel}=F_{g}\\sin \\theta\\,\\pm\\,F_{f}\\,\\pm F_{T}\\,\\pm\\,...=0$$","backgroundColorModified":false,"aid":null,"type":"$$","id":"7-0","font":{"family":"Calibri","color":"#000000","size":11},"ts":1634829288659,"cs":"BOx5GC1ASavRalJSKxedAA==","size":{"width":276,"height":20}}

  1. Block Moving with Constant Velocity on Incline
  • Again, the sum of all forces acting parallel to the incline’s surface is zero.

{"backgroundColor":"#ffffff","code":"$$\\sum_{}^{}F_{parallel}=F_{g}\\sin \\theta\\,\\pm\\,F_{f}\\,\\pm F_{T}\\,\\pm\\,...=0$$","backgroundColorModified":false,"aid":null,"type":"$$","id":"7-1","font":{"family":"Calibri","color":"#000000","size":11},"ts":1634829288659,"cs":"AMEbeqeu1YJtmlPK5+m4kg==","size":{"width":276,"height":20}}

  1. Block Accelerating along the Incline
  • The sum of all forces acting parallel to the incline’s surface must now follow Newton’s 2nd Law of Motion using the net acceleration of the object.

{"code":"$$\\sum_{}^{}F_{parallel}=F_{g}\\sin \\theta\\,\\pm\\,F_{f}\\,\\pm F_{T}\\,\\pm\\,...=ma_{net}$$","font":{"family":"Calibri","size":11,"color":"#000000"},"aid":null,"backgroundColorModified":false,"id":"7","backgroundColor":"#ffffff","type":"$$","ts":1634829328620,"cs":"e0q+wbdTVjvpVm6mbLPPKg==","size":{"width":305,"height":20}}

EXAMPLE 1 – Calculate the normal force provided by the incline.

  1. A 215-N box is placed on an inclined plane that makes a 35.0° angle with the horizontal.

EXAMPLE 2 – Solve the following for objects in equilibrium on an incline.

  1. A block of mass m = 2.0 kg is held in equilibrium on an incline of angle θ = 60.0° by a cable. Determine the force provided by the cable.
  1. A 10.0 kg block moves down a 30.0° incline at constant velocity, find the force of friction acting on the block.

EXAMPLE 3 – Solve the following for objects accelerating on an incline.

  1. A student moves a box of books up a frictionless hill angled 25.0° above the horizon. The student pushes with a force of 85 N parallel to the incline. The box has a mass of 35.0 kg. Find the acceleration of the box.
  1. A box of mass 10.0 kg slides down a 30.0° ramp with an acceleration of 1.20 m/s2. Determine the force of friction acting on the box.

AP Physics - Chapter 4

Lesson 22: Tension

Notes

LEARNING TARGET & JOURNALS

  • I can account for the effects of tension and contact forces.

NOTES

  • Tension is the measure of force acting on a string, cable, spring, etc.
  • Tension is most often found from another force that it is working against. http://cnx.org/resources/6b7d63bb053e61e7a14207745fce1d94/Figure_04_05_04.jpg
  • Tension created by two objects pulling on the same string follows Newton’s 3rd Law of Motion.                        

http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0008/incline_wpulleyFBD.gif

  • To find the tension in supporting cables of an object in equilibrium remember to sum the forces in both dimensions to zero.Image result for free body diagram tension
  • {"id":"8-0","backgroundColor":"#ffffff","font":{"size":"12","family":"Calibri","color":"#000000"},"code":"$$\\sum_{}^{}F_{x}=F_{1x}\\,+\\,F_{2x}\\,+\\,...=0$$","backgroundColorModified":false,"aid":null,"type":"$$","ts":1634829797072,"cs":"oU7uLK4fzuvj1BOVkl8yvg==","size":{"width":208,"height":21}}
  • {"aid":null,"backgroundColorModified":false,"type":"$$","code":"$$\\sum_{}^{}F_{y}=F_{1y}\\,+\\,F_{2y}\\,+\\,...=0$$","font":{"family":"Calibri","size":"12","color":"#000000"},"backgroundColor":"#ffffff","id":"8-1","ts":1634829824823,"cs":"OYjB5MZ9iffGLKKq1+dxTw==","size":{"width":204,"height":21}}

Two Blocks Against Each Other

  • Draw a free body diagram for each block separately and one for the total system of blocks together.
  • Use the free body diagram of the total system to find the net acceleration.
  • The acceleration for the total system is the same for each object individually.
  • Once the acceleration is found, then you can go back to each individual object to find any interaction forces or mass values.

{"code":"$$\\sum_{}^{}F=\\,\\left(\\sum_{}^{}m\\right)a_{net}$$","type":"$$","backgroundColorModified":false,"font":{"color":"#000000","size":11,"family":"Calibri"},"aid":null,"id":"9","backgroundColor":"#ffffff","ts":1634829862652,"cs":"rfrs4sJ8kfnH4IxSabgEbw==","size":{"width":138,"height":44}}

So to find the net acceleration using the following:

{"code":"$$a_{net}=\\,\\frac{\\Sigma F}{M+m}$$","aid":null,"backgroundColor":"#ffffff","font":{"color":"#000000","size":11,"family":"Calibri"},"type":"$$","id":"10","backgroundColorModified":false,"ts":1634829975063,"cs":"XNq6hzT50en3y1as16mBAQ==","size":{"width":96,"height":30}}

EXAMPLE 1 – Find the tension in the supporting cables that would hold the mass in equilibrium.

EXAMPLE 2 -         Three blocks are in contact with each other on a frictionless horizontal surface. A horizontal force, F, is applied to m1. For this problem, m1 = 2.00 kg, m2 = 3.00 kg, m3 = 4.00 kg, and F = 180 N to the right.

  1. Find the acceleration of the blocks.

  1. Find the resultant force on each block.
  1. Find the magnitude of the contact forces between the blocks.

AP Physics - Chapter 4

Lesson 23: Pulleys  and Tension

Notes

LEARNING TARGET & JOURNALS

  • I can incorporate the effects of tension forces redirected over pulleys.

NOTES

There are 3 common types of problems involving pulleys:

  1. Single Pulley with Two Masses (Atwood’s Machine)
  • Draw a free body diagram for each object individually and recognize the tension acting on both objects is the same in magnitude but opposite in direction, so it will cancel out of our calculations.
  • Determine if there is any net acceleration on the system to create the equation

{"type":"$$","aid":null,"id":"11","code":"$$\\sum_{}^{}F=\\,\\left(\\sum_{}^{}m\\right)a_{net}\\Rightarrow\\,\\left(M+m\\right)a_{net}=\\,Mg-mg$$","font":{"size":11,"family":"Calibri","color":"#000000"},"backgroundColor":"#ffffff","backgroundColorModified":false,"ts":1634830118499,"cs":"zrWaf+NQOhAZH3tHm2ZMQg==","size":{"width":332,"height":44}}

                

The above equation can find net acceleration:

{"backgroundColor":"#ffffff","aid":null,"code":"$$a_{net}=\\,\\frac{\\left(M-m\\right)}{\\left(M+m\\right)}g$$","id":"12-0","font":{"size":"10","color":"#000000","family":"Calibri"},"type":"$$","backgroundColorModified":false,"ts":1634830212890,"cs":"jQU/CoROulEnC49rk+Wp4Q==","size":{"width":105,"height":32}}

The acceleration can now be used with either object in a net force summation to solve for the tension (T) in the string

{"backgroundColorModified":false,"type":"$$","font":{"size":"10","family":"Calibri","color":"#000000"},"aid":null,"backgroundColor":"#ffffff","code":"$$Ma_{net}=\\,Mg-T$$","id":"13-0","ts":1634830237172,"cs":"U6DYBkGURyef7F5Hf1B81Q==","size":{"width":101,"height":12}}        or        {"font":{"family":"Calibri","color":"#000000","size":"10"},"code":"$$ma_{net}=\\,T-mg$$","type":"$$","backgroundColorModified":false,"backgroundColor":"#ffffff","aid":null,"id":"13-1","ts":1634830273672,"cs":"881b4ddlfMxxB4TRzx1VQg==","size":{"width":96,"height":12}}

  1. Block Over the Edge
  • The tension is all in the same string, albeit in two different dimensional directions
  • So, the two free-body diagrams can still be treated as a linear interaction of forces with the weight of the hanging object working against the friction acting on the other.
  • Establish a frame of reference based on the direction the hanging mass is pulling as positive.
  • If the surface is frictionless, then the weight of the hanging object will create the acceleration of the entire system of the sum of the two masses.
  • If the surface is not frictionless, the acceleration must be found by calculating the net force of the weight of the hanging mass and the friction acting on the mass on the surface.

                                

Frictionless Surface:

{"backgroundColorModified":false,"id":"14","aid":null,"font":{"family":"Calibri","size":"10","color":"#000000"},"type":"$$","code":"$$mg=\\,\\left(M+m\\right)a_{net}$$","backgroundColor":"#ffffff","ts":1634830323697,"cs":"Jnds9WVLfIEDAtVpWiQGuA==","size":{"width":112,"height":12}}

Solving for net acceleration

{"id":"12-1-0","font":{"color":"#000000","size":"10","family":"Calibri"},"code":"$$a_{net}=\\,\\frac{mg}{\\left(M+m\\right)}$$","backgroundColorModified":false,"aid":null,"backgroundColor":"#ffffff","type":"$$","ts":1634830405296,"cs":"LGhXB0y4cKJkdCAGl5B7Jg==","size":{"width":97,"height":28}}

Rough Surface:

{"backgroundColor":"#ffffff","font":{"family":"Calibri","color":"#000000","size":"10"},"type":"$$","aid":null,"backgroundColorModified":false,"id":"15","code":"$$\\sum_{}^{}F=\\left(M+m\\right)a_{net}=\\,mg-F_{f}$$","ts":1634830364796,"cs":"qNyO8rpKrFKPdTOnlpFTJA==","size":{"width":192,"height":18}}

Solving for net acceleration

{"code":"$$a_{net}=\\,\\frac{mg-F_{f}}{\\left(M+m\\right)}$$","aid":null,"backgroundColorModified":false,"id":"12-1","font":{"size":"10","color":"#000000","family":"Calibri"},"backgroundColor":"#ffffff","type":"$$","ts":1634830444091,"cs":"zEjImI6lzfsSH8xz6OhKJA==","size":{"width":97,"height":32}}

  1. Block over the edge of an inclineImage result for incline with pulley
  • Treat the same way as a block hanging over the edge of a flat table, EXCEPT
  • You must use the parallel component of gravity as the counterforce to the hanging mass.

{"id":"16","font":{"color":"#000000","size":11,"family":"Calibri"},"code":"$$\\sum_{}^{}F=\\left(M+m\\right)a_{net}=\\,Mg-mg\\sin \\theta\\,-\\,F_{f}$$","backgroundColorModified":false,"backgroundColor":"#ffffff","type":"$$","aid":null,"ts":1634831398427,"cs":"2pM6vm5RGgmfxhW5vhzX0g==","size":{"width":288,"height":20}}

  • To find the tension, use Newton’s 2nd Law for either mass, one at a time.

{"aid":null,"type":"$$","code":"$$Ma_{net}=\\,Mg-T$$","id":"17","backgroundColorModified":false,"font":{"size":"11","color":"#000000","family":"Calibri"},"backgroundColor":"#ffffff","ts":1634831425072,"cs":"AO7SZ9eHM3sP1fVqSP9x+g==","size":{"width":112,"height":12}}  or        {"id":"18","font":{"color":"#000000","family":"Calibri","size":"11"},"backgroundColorModified":false,"backgroundColor":"#ffffff","aid":null,"type":"$$","code":"$$ma_{net}=\\,T-mg\\sin \\theta$$","ts":1634831467280,"cs":"PVjjEiYR8xgaLAxglrwT2Q==","size":{"width":136,"height":12}}

EXAMPLE 1 – A 2.0-kg mass (mA) and a 3.0-kg mass (mB) are attached to a lightweight cord that passes over a massless-frictionless pulley. Find the acceleration of the smaller mass.

EXAMPLE 2 – Solve the following problems with a block sliding across the table. Assume the table is frictionless unless told otherwise. Also, assume the pulley is massless and frictionless.

  1. A block with mass m1 = 4.00 kg and a ball with mass m2 = 7.00 kg are connected by a light string that passes over a massless-frictionless pulley. Find the acceleration of the two objects and the tension in the string.

  1. Two packing crates of masses 10.0 kg and 5.00 kg are connected by a light string that passes over a massless-frictionless pulley. The 5.00-kg crate lies on a smooth incline of angle 40.0°. Find the acceleration of the 5.00-kg crate and the tension in the string.