Q: A 50 kg person steps off a diving platform that is 10 meters above the water. What is their speed just before hitting the water? (Ignore air resistance.)

$$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 10} \approx 14 m/s$$

Q: You push a 10 kg box up a 30 degree hill with a force of 85 N directed uphill (parallel to the hill's surface). The coefficient of kinetic friction between the box and the hill is 0.1. What is the magnitude of acceleration of the box?

$$a = \frac{F - mg \sin \theta - \mu mg \cos \theta}{m} = \frac{85 - 10 \times 9.8 \times 0.5 - 0.1 \times 10 \times 9.8 \times 0.866}{10} \approx 2.75 m/s$$²

Question 1

A car traveling at 30 m/s begins to decelerate at a constant rate of 5 m/s². Which equation can be used by itself to find the stopping distance of the car?

Accepted Answer

$$v_f^2$ = v_i^2$ + 2a(x_f - x_i)$$

Question 2

A 50 kg person steps off a diving platform that is 10 meters above the water. What is their speed just before hitting the water? (Ignore air resistance.)

Accepted Answer

$$v = \sqrt{2gh} = \sqrt{2 	imes 9.8 	imes 10} \approx 14 m/s$$

Question 3

Which of the following will result in more acceleration when pulling a box with a rope?

Accepted Answer

Decreasing the mass of the box

Question 4

A 10 kg box experiences a force of 15 N in the positive x-direction and 8 N in the negative x-direction. What is the magnitude of the net force acting on the box?

Accepted Answer

$$F_{net} = 15 - 8 = 7 N$$

Question 5

A 10 kg box is being lifted straight upwards with an acceleration of 2 m/s² by a rope. Which of the following forces are acting on the box?

Accepted Answer

Tension and gravity

Question 6

A ball is being swung in a vertical circle at a uniform speed by a rope. When is the tension in the rope the greatest?

Accepted Answer

When the ball is at the bottom

Question 7

A 500 kg roller coaster cart travels at a constant speed in a horizontal circular loop of radius 10 meters. If it experiences a net force of 5000 Newtons, how fast is it moving?

Accepted Answer

$$F = \frac{mv^2$}{r} \Rightarrow v = \sqrt{\frac{Fr}{m}} = \sqrt{\frac{5000 	imes 10}{500}} = 10 m/s$$

Question 8

Which statement about torque is correct?

Accepted Answer

A nonzero torque always gets bigger in magnitude if the associated force gets bigger in magnitude.

Question 9

Which of these are units of power?

Accepted Answer

W (Watt)

Question 10

Which of the following can be used to calculate the work done by a conservative force?

Accepted Answer

$$W = \Delta PE$$

Question 11

A box is initially pressed against a compressed spring. The spring is then released, launching the box up a ramp with friction. Which term in the energy conservation equation is NOT zero?

Accepted Answer

$$PE_{spring}$$

Question 12

An engine has a power output of 300 W when a car is going at a constant 10 m/s. What is the magnitude of the net force resisting the motion of the car?

Accepted Answer

$$F = \frac{P}{v} = \frac{300}{10} = 30 N$$

Question 13

You push a 10 kg box up a 30 degree hill with a force of 85 N directed uphill (parallel to the hill's surface). The coefficient of kinetic friction between the box and the hill is 0.1. What is the magnitude of acceleration of the box?

Accepted Answer

$$a = \frac{F - mg \sin 	heta - \mu mg \cos 	heta}{m} = \frac{85 - 10 	imes 9.8 	imes 0.5 - 0.1 	imes 10 	imes 9.8 	imes 0.866}{10} \approx 2.75 m/s$$²

Question 14

An 80-cm uniform 10 kg bar is resting on two scales, one at either end. A smaller 4-kg mass is placed at a distance of 20 cm from the left end. Which scale reads a higher weight, and what is the difference between the weights read by the two scales?

Accepted Answer

The left scale reads higher; the difference is $$\Delta W = 4 	imes 9.8 	imes \frac{60}{80} = 29.4 N$$

Question 15

A 2 kg block is attached to a spring ($$k = 400 N/m) $$and oscillates back and forth without friction. When the spring is compressed 0.1 m from equilibrium, the block has a speed of 2 m/s. What is the fastest speed the block achieves during the bouncing?

Accepted Answer

$$v_{max} = \sqrt{v^2$ + \frac{kx^2$}{m}} = \sqrt{2^2$ + \frac{400 	imes 0.1^2$}{2}} = \sqrt{4 + 2} = \sqrt{6} \approx 2.45 m/s$$

Fundamental Concepts in College Physics: Kinematics, Dynamics, Work, Energy, and Rotational Motion

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