Practice Test Solutions | |

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## Solutions

**MULTIPLE CHOICE ANSWERS:**

1. ANS: C

Calculate the cosine of F to find the horizontal component of the force of the right. Add the component force to the force on the left to calculate the net horizontal force on the block. Then use F-ma to calculate the acceleration.

2. ANS: B

Draw a free-body diagram to help you with this question. The weight of the object is suppsed to be 39.2N, which means that the elevator is accelerating upwards. The net force on the block is the Tension – Weight. Divide that by the mass of the block will yield the acceleration of the elevator.

3. ANS: D

Find the component of the force of gravity parallel to the incline and subtract that by the upward pushing force. Divide this net force by the mass of the block.

4. ANS: C

Calculate the change in velocity and divide the change by the change in time. This yields the acceleration of the object. Using Newton’s 2nd Law, F=ma, find F.

5. ANS: D

Find the horizontal component of the northwest force and add that to the east force to find the net horizontal component of the resultant force. Calculate the vertical component of the northwest force and that yields the vertical component of the resultant force. Find the magnitude of the resultant force using the Pythagorean theorem. Then, using F=ma, find a.

6. ANS: D

The least force is used when the force is applied parallel to the direction of motion. This happens when force applied is divided by the normal force, which yields the coefficient of friction. This is also the definition of cot q.

7. ANS: D

The amount of net force on the two blocks is the weight of the two blocks minus the force caused by the acceleration of the elevator. Since the net force is given and the unknown mass is the only unknown variable, calculus m using F=ma.

8. ANS: C

Find the horizontal component of the force then subtract the friction force from that. Using F=ma, find acceleration.

9. ANS: D

Calculate the individual net force on each box by subtracting the friction force from the applied force. Then subtract the individual net forces.

1. Since Fx = mgsinθ and fs = μsmgcosθ, the total force you have to overcome is

F = Fx + fs = mgsinθ + μsmgcosθ.

Substituting values gives us

F = (126 N) + (7.5 N) = 133 N

Note that most of this force is due to the component of the weight along the ramp.

2. a. Since the total weight of the system is 480.N, the total mass of the system = 480. N / 9.81 m/s2 = 49.0 kg

Taking upward as positive, the acceleration of the system is found using the second law.

∑Fy = 2T - mg = may

Solving for ay,

ay = [(250. N + 250. N) - 480. N] / 49.0 kg

ay = + 0.408 m/s^2

(positive sign indicates upward)

2. b.The downward force the child exerts on the chair has the same magnitude

as the upward normal force exerted on the child by the chair.

Applying the second law,

∑Fy = T + Fn - mchildg = mchilday

Fn = mchilday + mchildg - T

Fn = (320. N / 9.81 m/s^2)(0.408 m/s^2) + (320. N) - (250. N)

Fn = 83.3 N

3. The normal force exerted on the block by the incline is

Fn = mg cos(15.0) = (19.6 N)cos(15.0) = 18.9 N

The friction force is

f = μkFn = (0.250)(18.9 N) = 4.73 N

We choose up the incline as the positive x direction.

The net force parallel to the incline while the block moves up the incline is

∑Fx = -mg sin(15.0) - f = -9.81 N

The acceleration of the block is

ax = ∑Fx / m = (-9.81 N) / (2.00 kg) = -4.90 m/s^2

The distance the block travels up the incline before stopping is found using

vf^2 = vi^2 + 2ad

0 = (2.50 m/s)2 + 2(-4.90 m/s2)d

d = 0.637 m

As the block starts from rest and comes back down the incline, the net force is

∑Fx = -mg sin(15.0) + f = -0.340 N

and the acceleration is

ax = ∑Fx / m = (-9.81 N) / (2.00 kg) = -0.170 m/s^2

The speed of the block when it returns to the bottom of the incline is found using

vf^2 = vi^2 + 2ad

vf^2 = 2(-0.170 m/s^2)(-0.637 m)

vf = 0.465 m/s

F = Fx + fs = mgsinθ + μsmgcosθ.

Substituting values gives us

F = (126 N) + (7.5 N) = 133 N

Note that most of this force is due to the component of the weight along the ramp.

2. a. Since the total weight of the system is 480.N, the total mass of the system = 480. N / 9.81 m/s2 = 49.0 kg

Taking upward as positive, the acceleration of the system is found using the second law.

∑Fy = 2T - mg = may

Solving for ay,

ay = [(250. N + 250. N) - 480. N] / 49.0 kg

ay = + 0.408 m/s^2

(positive sign indicates upward)

2. b.The downward force the child exerts on the chair has the same magnitude

as the upward normal force exerted on the child by the chair.

Applying the second law,

∑Fy = T + Fn - mchildg = mchilday

Fn = mchilday + mchildg - T

Fn = (320. N / 9.81 m/s^2)(0.408 m/s^2) + (320. N) - (250. N)

Fn = 83.3 N

3. The normal force exerted on the block by the incline is

Fn = mg cos(15.0) = (19.6 N)cos(15.0) = 18.9 N

The friction force is

f = μkFn = (0.250)(18.9 N) = 4.73 N

We choose up the incline as the positive x direction.

The net force parallel to the incline while the block moves up the incline is

∑Fx = -mg sin(15.0) - f = -9.81 N

The acceleration of the block is

ax = ∑Fx / m = (-9.81 N) / (2.00 kg) = -4.90 m/s^2

The distance the block travels up the incline before stopping is found using

vf^2 = vi^2 + 2ad

0 = (2.50 m/s)2 + 2(-4.90 m/s2)d

d = 0.637 m

As the block starts from rest and comes back down the incline, the net force is

∑Fx = -mg sin(15.0) + f = -0.340 N

and the acceleration is

ax = ∑Fx / m = (-9.81 N) / (2.00 kg) = -0.170 m/s^2

The speed of the block when it returns to the bottom of the incline is found using

vf^2 = vi^2 + 2ad

vf^2 = 2(-0.170 m/s^2)(-0.637 m)

vf = 0.465 m/s