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Exercise 2-2 Solution


1.  Which of Newton's laws best explains why motorists should buckle-up?


     A) the first law     Answer: A

     B) the second law

     C) the third law

     D) the law of gravitation


Simole illustration of the law


2.  A rocket moves through empty space in a straight line with constant speed.  It is far from the

    gravitational effect of any star or planet.  Under these conditions, the force that must be applied to

    the rocket in order to sustain its motion is


     A) equal to its weight.

     B) equal to its mass.

     C) dependent on how fast it is moving.

     D) zero.     Answer: D


Newton's first law


3.  You are standing in a moving bus, facing forward, and you suddenly fall forward as the bus comes

     to an immediate stop.  What force caused you to fall forward?


     A) gravity

     B) normal force due to your contact with the floor of the bus

     C) force due to friction between you and the floor of the bus

     D) There is not a force leading to your fall.     Answer: D


It is no longer an inertial reference frame (one that is in motion with constant velocity or one that is not in motion). Newton's laws hold only

in inertial reference frames.


4.  A constant net force acts on an object.  Describe the motion of the object.


     A) constant acceleration     Answer: A

     B) constant speed

     C) constant velocity

     D) increasing acceleration


F = ma  If F is constant (and m is not changing), then a must be constant

If the object is dropped from rest, then the constant force is the force of gravity

in this case v = at and, with constant a (g, acceleration of gravity), v changes with time

If the object is sliding along a table (for example), then the contant net force (considering friction) results in a constant a also.

Remember the definition of the unit of force (Newton).

A Newton is the amount of force required to give 1 kg an acceleration of 1 m per second squared.


5.  A net force F accelerates a mass m with an acceleration a.  If the same net force is applied to

     mass 2m, then the acceleration will be


     A) 4a.    

     B) 2a.

     C) a/2.     Answer: C

     D) a/4.


F = ma so a = F/m 

If m is doubled, then a is divided by 2


6.  If you blow up a balloon, and then release it, the balloon will fly away.  This is an illustration of


     A) Newton's first law.

     B) Newton's second law.

     C) Newton's third law.      Answer: C

     D) Galileo's law of inertia.


Simple illustration of the law


7.  A 20-ton truck collides with a 1500-lb car and causes a lot of damage to the car.  Since a lot of

     damage is done on the car


     A) the force on the truck is greater than the force on the car.

     B) the force on the truck is equal to the force on the car.      Answer: B

     C) the force on the truck is smaller than the force on the car.

     D) the truck did not slow down during the collision.


The force is the same by Newton's third law. Since F = ma, the one with the smaller mass attains a higher acceleration with the impact, resulting in more damage.


8.  A golf club hits a golf ball with a force of 2400 N.  The golf ball hits the club with a force


     A) slightly less than 2400 N.

     B) exactly 2400 N.        Answer: B

     C) slightly more than 2400 N.

     D) close to 0 N.


Same as problem 8


9. Mass and weight


      A) both measure the same thing.

      B) are exactly equal.

      C) are two different quantities.      Answer: C

      D) are both measured in kilograms.


Mass is a measure of the inertia (resistance to a change in motion) of an object. Weight is a force, the pull of gravity acting on an object.


10. A stone is thrown straight up.  At the top of its path, the net force acting on it is


      A) greater than its weight.

      B) greater than zero, but less than its weight.

      C) instantaneously equal to zero.

      D) equal to its weight.     Answer: D


The force acting on it is mg which is its weight, See problem 9


11. A 20-N weight and a 5.0-N weight are dropped simultaneously from the same height.  Ignore air

      resistance.  Compare their accelerations.


      A) The 20 N weight accelerates faster because it is heavier.

      B) The 20 N weight accelerates faster because it has more inertia.

      C) The 5.0 N weight accelerates faster because it has a smaller mass.

      D) They both accelerate at the same rate because they have the same weight to mass ratio.     Answer: D


F = ma  20 = m1a     5 = m2a    since the accelerations are the same, 20/m1 = 5/m2    

Saying they have the same weight to mass ratio is synonomous with saying they have the same acceleration - which is true


12. An object of mass m is hanging by a string from the ceiling of an elevator.  The elevator is moving

      up at constant speed.  What is the tension in the string?


      A) less than mg

      B) exactly mg      Answer: B

      C) greater than mg

      D) cannot be determined without knowing the speed


WIth constant speed, it is an inertial reference frame. The only force acting on the mass is due to gravity


13. A block of mass M slides down a frictionless plane inclined at an angle θ with the horizontal.  The

      normal reaction force exerted by the plane on the block is


      A) Mg.

      B) Mg sin θ.

      C) Mg cos θ.      Answer: C

      D) zero, since the plane is frictionless.


Draw a free-body diagram to see this


14. A block of mass M slides down a frictionless plane inclined at an angle θ with the horizontal.  The

      gravitational force is directed  


      A) parallel to the plane in the same direction as the movement of the block.

      B) parallel to the plane in the opposite direction as the movement of the block

      C) perpendicular to the plane.

      D) toward the center of the Earth.      Answer: D


Regardless of the reference frame used, this is always true 

15. Two toy cars (16 kg and 2.0 kg) are released simultaneously on an inclined plane that makes an

      angle of 30° with the horizontal.  Make a statement which best describes their acceleration after

      being released.


      A) The 16-kg car accelerates 8 times faster than the 2.0-kg car.

      B) The 2.0-kg car accelerates 8 times faster than the 16-kg car.

      C) Both cars accelerate at the same rate.     Answer: C

      D) none of the above


Galileo showed this with his experiments - which we duplicated in a lab


16. An object sits on a frictionless surface.  A 16-N force is applied to the object, and it accelerates at

      2.0 m/s2.  What is the mass of the object?


      A) 4.0 kg

      B) 8.0 kg     Answer: B

      C) 32 kg

      D) 78 N



17. Starting from rest, a 4.0-kg body reaches a speed of 8.0 m/s in 2.0 s.  What is the net force acting

      on the body?


      A) 4.0 N

      B) 8.0 N

      C) 16 N       Answer: C

      D) 32 N


a = change in speed/change in time = (8 - 0)/(2 - 0) = 4

F = ma = 4x4 = 16 

18. Sue and Sean are having a tug-of-war by pulling on opposite ends of a 5.0-kg rope.  Sue pulls

      with a 15-N force.  What is Sean's force if the rope accelerates toward Sue at 2.0 m/s2?


      A) 3.0 N

      B) 5.0 N     Answer: B

      C) 25 N

      D) 50 N


It accelerates towards Sue, so her force must exceed that of Sean's

FSue - FSean = mropearope

15 - FSean = 5x2   = 10 gives FSean = 15 - 10 = 5


19. A traffic light of weight 100 N is supported by two ropes as shown in the

       figure on the right. What are the tensions in the ropes?


      A) 50 N

      B) 63 N

      C) 66 N

      D) 83 N     Answer: D



Sum forces in y direction and obtain FTy = 50 N

FT = FTy/sin37 = 50/0.602 = 83 N


20. Two boxes of masses m and 2m are in contact with each other on a

       frictionless surface.  (See figure on the right). What is the acceleration of the

       more massive box?


      A) F/m

      B) F/(2m)

      C) F/(3m)     Answer: C

      D) F/(4m)

F = ma gives a = F/3M


21. A 10-kg box sitting on a horizontal surface is pulled by a 5.0-N force.  A 3.0-N friction force retards

      the motion.  What is the acceleration of the object?


      A) 0.20 m/s2     Answer: A

      B) 0.30 m/s2 

      C) 0.50 m/s2 

      D) 5.0 m/s2 


Sum forces in x direction, resultant force F = 2 N = ma = 10a gives a = 0.2


22. A horizontal force of 5.0 N accelerates a 4.0-kg mass, from rest, at a rate of 0.50 m/s2 in the

      positive direction.  What friction force acts on the mass?


      A) 2.0 N

      B) 3.0 N     Answer: B

      C) 4.0 N

      D) 5.0 N



23. A 10-kg mass slides down a flat hill that makes an angle of 10° with the horizontal.  If friction is

      negligible, what is the resultant force on the sled?   NOTE:  sin10 degrees = 0.174       cos 10 degrees = 0.985


      A) 1.7 N

      B) 17 N     Answer: B

      C) 97 N

      D) 98 N


Draw a free body diagram and find the component of the gravitational force in the x direction


24. A mass is placed on a smooth inclined plane with an angle of 37° to the horizontal.  If the inclined

      plane is 5.0-m long, how long does it take for the mass to reach the bottom of the inclined plane

      after it is released from rest?     NOTE:  sin 37 degrees = 0.602       cos 37 degrees = 0.799


      A) 1.3 s     Answer: A

      B) 1.2 s

      C) 1.1 s

      D) 1.0 s


Find F in x direction using angles, then set F = ma, find that abox = 5.9 m/s2

use x = 1/2 at2 and solve for t


25. An object with a mass m slides down a rough 37° inclined plane where the coefficient of kinetic

      friction is 0.20.  What is the acceleration of the object?


      A) 4.3 m/s2     Answer: A

      B) 5.9 m/s2 

      C) 6.6 m/s2 

      D) 7.8 m/s2 


Find F of gravity in rotated x direction and subtract from it Ffr. Set  equal to mabox


26. An object with a mass m slides down a rough 37° inclined plane where the coefficient of kinetic

      friction is 0.20.  If the plane is 10 m long and the mass starts from rest, what will be its speed at

      the bottom of the plane?


      A) 12 m/s

      B) 11 m/s

      C) 9.7 m/s

      D) 9.3 m/s     Answer: D


F of gravity in x direction = mgsin 37 = FP

Normal force = mgcos 37

FFr = mgcos 37x0.20

Sum forces in x direction and set equal to mabox

mgsin37 - mg cos 37x0.2 = mabox

Solve for acceleration of the box 

x = 1/2at2 so solve for t       use v = at to solve for v


27. A bulldozer drags a log weighing 500 N along a rough surface.  The cable attached to the log

      makes an angle of 30.0° with the ground.  The coefficient of static friction between the log and the

      ground is 0.500.  What minimum tension is required in the cable in order for the log to begin to



      A) 224 N      Answer: A

      B) 268 N      Answer B

      C) 289 N    

      D) 500 N


Sum forces in y direction (positive is up): Fsin30 + FN - 500 = 0 (because no acceleration in y direction)

FN = 500 - Fsin30 = 500 - 0.5F                

FN = 500 - 0.5F                

FFr = 0.5(500 - 0.5F) = 250 - 0.25F

To just move FFr < F

FFr = F  

gives 250 - 0.25F = Fcos 30

250 - 0.25 F = 0.866 F

1.116F = 250 gives F = 224

So, anything over 224