Physical Science 9 Practice Test for Forces Solutions
 

1. State any two of Newton's Laws in your own words and explain them using examples.

N1:  If an object is experiencing an acceleration (or change in velocity, or change in speed and/or direction) it must have a net force on it.  Likewise, if it does not have a net force on it, it will not accelerate.

Example: Consider the moon orbiting in a circle around the earth at constant speed.  Though its speed is not changing its direction is, so it is accelerating, hence it must have a net force on it.

N2:  An object of mass m will experience an acceleration a when a net force F_net is applied to it, such that:

F_net= ma  (the net force points in the same direction as the acceleration).

Example: If I (of mass 75 kg) were in outer space, away from any planet or star, and was being pulled by a rope with tension 100N to the right and a rope of tension 250N to the left I would feele a net force to the left of 150N.  From N2 I can calculate my acceleration algebraically as:

a = F_net/m

in this case it will give:

a = 150N left / 75 kg = 2.0 m/s² left

N3: If object 1 feels a force from object 2 (the action) then object 2 will experience the same size force from object 1 but in the opposite direction (the reaction).

Example: If I am standing on the ground and the ground pushes up on me with a normal force of 735N then I will push down on the ground with a normal force of 735N.

 

2. Consider a 0.80 kg brick sitting on the ground. Draw the free body diagram for the brick labeling it with the types of forces and their values. Describe the reaction forces here.

The brick will have two arrows of force: an arrow pointing down from its center labeled “W” or “weight” or “force of gravity” which will be equal to W = mg = 0.80kg×9.8 m/s² = 7.84 N and an arrow pointing up from the bottom of the brick labeled “N” or “normal force”.  The normal force must be equal to the weight since the brick is not moving so it has zero acceleration (from N1 and/or N2).  So, you label the normal force with 7.84N as well.

The reaction forces will be:

To the normal force:  the brick will push down on the ground with 7.84 N.

To the weight: the brick will pull up on the earth with a gravity force of 7.84 N.

3. Draw a free body diagram for a ball rolling down a hill naming the forces operating. Include friction.

There will be a normal force directed from the surface of contact between the ball and the ground point out at right angles from the hill. 

There will be a frictional force from the surface of contact backward along the hill.

There will be a weight force from the center of the ball directed down.

4. Consider a 2000 kg car being towed along a flat road with constant speed and direction. If the cable towing the car is also horizontal and has a tension force of 200N, what will the force of friction be?

Since it has constant speed and direction it has no acceleration and hence no net force.  The 200N towing the car must be balanced by 200N of friction resisting the pull.

If the cars hits a patch of oil and the force of friction reduces to 100N, what would the acceleration of the car be?

Now the net force will be 100N in the direction it is being towed (200N towing force minus 100N friction force).  From N2:

a = F_net/m

= 100N/2000kg = 0.050 m/s² in the direction of towing.

(Bonus: if the car is traveling at 10m/s when it hits the oil patch, how fast will it be traveling 2.0 s later?)

From last unit the equation for acceleration is:

a = Δv/Δt

so Δv = a Δt

Here we have a=0.050 m/s² and Δt = 2.0s from the question so:

Δv = 0.050 m/s²2.0s = 0.10m/s in the direction of towing.

If it starts at 10m/s in the direction of towing and GAINS 0.10m/s it will have 10.1m/s in the direction of towing.

5. Create your own scenario. Draw a free body diagram and label it with force types and values. Calculate the acceleration of the object. Give two reaction forces.

 

It’s up to you!