Question 1

A car is traveling at 7.0 m/s when the driver applies the brakes. The car moves 1.5 m before it comes to a complete stop. If the car had been moving at 14 m/s, how far would it have continued to move after the brakes were applied? Assume the braking force in both cases is constant and the same.

Accepted Answer

The stopping distance is proportional to the square of the speed. Doubling the speed from 7.0 m/s to 14 m/s (which is twice) means the stopping distance will be four times greater (since 2^2 = 4). Therefore, 1.5 m * 4 = 6.0 m.

Question 2

A 1.0-kg ball on the end of a string is whirled at a constant speed of 2.0 m/s in a horizontal circle of radius 1.5 m. What is the work done by the centripetal force during one revolution?

Accepted Answer

The work done by a centripetal force in one revolution is always zero because the force is always perpendicular to the displacement. Therefore, the dot product of force and displacement will be zero.

Question 3

Which one of the following statements concerning kinetic energy is true?

Accepted Answer

Kinetic energy is always positive because it is calculated as 1/2 mv^2, where m is mass and v is velocity. Since mass and the square of velocity are always positive, kinetic energy cannot be negative.

Question 4

In which one of the following situations will there be an increase in kinetic energy?

Accepted Answer

As the ball rolls downhill, it gains kinetic energy due to its increasing speed. A projectile at its maximum height has zero velocity, a box pulled at a constant speed has constant kinetic energy, a satellite in circular orbit has constant kinetic energy, and a stone whirled in a horizontal circle has constant kinetic energy because its speed is constant.

Question 5

Which one of the following situations is an example of an object with a non-zero kinetic energy?

Accepted Answer

A satellite in geosynchronous orbit is moving at a constant speed around the Earth, thus it has non-zero kinetic energy due to its motion.

Question 6

Which one of the following has the largest kinetic energy?

Accepted Answer

The Earth moving in its orbit around the sun has the largest kinetic energy due to its massive size and high velocity compared to the other options.

Question 7

An experimental 1500-kg car travels at a constant speed of 22 m/s around a circular test track that is 80 m across. What is the kinetic energy of the car?

Accepted Answer

Kinetic energy = 1/2 * mv^2 = 1/2 * 1500 kg * (22 m/s)^2 = 3.6 × 10^5 J

Question 8

A concrete block is pulled 7.0 m across a frictionless surface by means of a rope. The tension in the rope is 40 N; and the net work done on the block is 247 J. What angle does the rope make with the horizontal?

Accepted Answer

The answer of A concrete block is pulled 7.0 m...

Question 9

Use the work-energy theorem to find the force required to accelerate an electron (m = 9.11 × 10^-³¹ kg) moving along the x axis from 4.00 × 10⁶ m/s to 1.60 × 10⁷ m/s in a distance of 0.0125 m.

Accepted Answer

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. The change in kinetic energy is calculated using the formula ΔKE = 0.5 * m * (v_f^2 - v_i^2). The work done is also equal to the force times the distance (W = F * d). Solving for force, F = ΔKE / d. Substituting the given values, the force required is approximately 8.75 × 10^-15 N.

Question 10

Mike is cutting the grass using a human-powered lawn mower. He pushes the mower with a force of 45 N directed at an angle of 41° below the horizontal direction. Calculate the work that Mike does on the mower each time he pushes it 9.1 m across the yard.

Accepted Answer

The answer of Mike is cutting the grass using a...

Question 11

The kinetic energy of an 1800-kg truck is 7.2 × 10⁵ J. What is the speed of the truck?

Accepted Answer

The formula for kinetic energy is: 
KE = 1/2mv^2 
where KE is kinetic energy, m is mass, and v is velocity.

Rearranging this formula, we get: 
v = √(2KE/m)

Plugging in the values, we get: 
v = √(2(7.2 × 10^15 J)/(1800 kg)) ≈ 28 m/s

Therefore, the correct answer is option C: 28 m/s.

Question 12

A 10.0-g bullet traveling horizontally at 755 m/s strikes a stationary target and stops after penetrating 14.5 cm into the target. What is the average force of the target on the bullet?

Accepted Answer

The average force of the target on the bullet is equal to the change in momentum of the bullet divided by the time of impact. The change in momentum is equal to the mass of the bullet times its final velocity minus its initial velocity. The time of impact is equal to the distance traveled by the bullet divided by its average velocity. The average velocity is equal to the sum of the initial and final velocities divided by 2.Plugging in the given values, we get:$$	ext{Average force}=\frac{m(v_f-v_i)}{t}=\frac{(0.0100	ext{ kg})(-755	ext{ m/s}-0	ext{ m/s})}{(14.5	imes10^{-2}	ext{ m})/(0.5(755	ext{ m/s}+0	ext{ m/s})}$$$$=\boxed{2.07	imes10^5	ext{ N}}$$

Question 13

A force with a magnitude of 25 N and directed at an angle of 37° above the horizontal is used to move a 10-kg crate along a horizontal surface at constant velocity. How much work is done by this force in moving the crate a distance of 15 m?

Accepted Answer

The work done by the force is calculated using the formula W = F * d * cos(θ), where F is the force magnitude, d is the distance, and θ is the angle. Here, W = 25 N * 15 m * cos(37°) = 300 J.

Question 14

A constant force of 25 N is applied as shown to a block which undergoes a displacement of 7.5 m to the right along a frictionless surface while the force acts. What is the work done by the force?

Accepted Answer

The answer of A constant force of 25 N is...

Question 15

How much energy is dissipated in braking a 1200-kg car to a stop from an initial speed of 30 m/s?

Accepted Answer

The kinetic energy of the car when it is traveling at 30 m/s is given by KE = 1/2mv^2 = 1/2(1200 kg)(30 m/s)^2 = 540,000 J. This is the amount of energy that needs to be dissipated in order to bring the car to a stop. Therefore, the answer is C) 540,000 J.

Question 16

Julie carries an 8.0-kg suitcase as she walks 18 m along a horizontal walkway to her hotel room at a constant speed of 1.5 m/s. How much work does Julie do in carrying her suitcase?

Accepted Answer

The work done is zero because the force Julie applies is vertical (to counteract gravity) and the displacement is horizontal, so the angle between the force and displacement is 90 degrees, resulting in no work done.

Question 17

In which one of the following situations is zero net work done?

Accepted Answer

If a box is pulled across a rough floor at constant velocity, the net work done on the box is zero because the force of friction acting on the box is equal and opposite to the force applied to move the box. This means that the work done by the applied force is canceled out by the negative work done by the frictional force, resulting in zero net work.

Question 18

A 5.00-kg block of ice is sliding across a frozen pond at 2.00 m/s. A 7.60-N force is applied in the direction of motion. After the ice block slides 15.0 m, the force is removed. The work done by the applied force is

Accepted Answer

The work done by a force is calculated using the formula: Work = Force × Distance × cos(θ). Here, the force is 7.60 N, the distance is 15.0 m, and the force is in the direction of motion (θ = 0 degrees, cos(0) = 1). Therefore, Work = 7.60 N × 15.0 m × 1 = 114 J. Since the force is in the direction of motion, the work done is positive, so the answer is +114 J.

Question 19

Complete the following statement: Work may be expressed using all of the following units except

Accepted Answer

Work is a measure of energy transfer and is typically expressed in units of energy such as joules, ergs, or foot-pounds. A watt is a unit of power, not work.

Question 20

The kinetic energy of a car is 8 × 10⁶ J as it travels along a horizontal road. How much work is required to stop the car in 10 s?

Accepted Answer

The work required to stop the car is equal to its initial kinetic energy, which is 8 × 10⁶ J.

Quiz 6: Work and Energy