Question 1

Compared to yesterday, you did 3 times the work in one-third the time. To do so, your power output must have been

Accepted Answer

Power is work done per unit time. If you did 3 times the work in one-third the time, your power output is 3 / (1/3) = 9 times yesterday's power output.

Question 2

FIGURE 7-1 
-Which of the graphs in Figure 7-1 represents a spring which gets less stiff the more it is stretched?

Accepted Answer

The answer of FIGURE 7-1 
-Which of the graphs in...

Question 3

A truck has four times the mass of a car and is moving with twice the speed of the car. If K_t and K_c refer to the kinetic energies of truck and car respectively, it is correct to say that

Accepted Answer

The answer of A truck has four times the mass...

Question 4

An object hits a wall and bounces back with half of its original speed. What is the ratio of the final kinetic energy to the initial kinetic energy?

Accepted Answer

The kinetic energy of an object is proportional to the square of its speed. If the object bounces back with half of its original speed, its final speed will be 1/2^1/2 times its initial speed. Thus, its final kinetic energy will be proportional to its final speed squared, which is (1/2^1/2)^2 = 1/2. Therefore, the ratio of the final kinetic energy to the initial kinetic energy is 1/2 divided by 1, which is 1/2 or 0.5. Answer B is the only option that matches this result.

Question 5

If the net work done on an object is negative, then the object's kinetic energy

Accepted Answer

According to the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy. If the net work is negative, it means that work was done against the motion of the object, leading to a decrease in its kinetic energy.

Question 6

FIGURE 7-1 
-Which of the graphs in Figure 7-1 illustrates Hooke's Law?

Accepted Answer

The answer of FIGURE 7-1 
-Which of the graphs in...

Question 7

In the SI system of units, power has the same units as

Accepted Answer

Power is measured in watts (W), which is equivalent to joules per second (J/s). Since 1 watt is also equal to 1 newton meter per second (Nm/s), option C is correct.

Question 8

As compared to Jack, Jill does twice the work in half the time. Jill's power output is

Accepted Answer

Power is defined as work done per unit time. Since Jill does twice the work in half the time compared to Jack, her power output is four times Jack's power output.

Question 9

If the net work done on an object is zero, then the object's kinetic energy

Accepted Answer

If the net work done on an object is zero, then the object's kinetic energy will remain the same. This is because work done on an object is directly related to the change in kinetic energy of the object, as stated by the work-energy theorem: Wnet = ΔKE. If Wnet is zero, then the change in kinetic energy (ΔKE) is zero as well, which means the kinetic energy of the object remains the same.

Question 10

On a force vs. position graph, the area under curve is a representation of

Accepted Answer

The area under a force vs. position graph represents the work done by the force. Work is defined as the product of force and displacement in the direction of the force. Therefore, the area under the curve on the graph represents the work done by the force.

Question 11

The ratio of the final kinetic energy to the initial kinetic energy of an object is one half. If the initial velocity of the object is 10 m/s, what is the final velocity?

Accepted Answer

The ratio of final kinetic energy to initial kinetic energy is 1/2. The formula for kinetic energy is KE = (1/2)mv^2, where m is the mass of the object and v is its velocity. Since the mass is constant, we can simplify the ratio to (1/2)(vf^2)/(vi^2) = 1/2. Solving for vf, we get vf = sqrt(vi^2/2) = sqrt(100/2) = 7.1 m/s. Therefore, the final velocity is 7.1 m/s.

Question 12

John's mass is half the mass of Jill. They both start walking and John moves twice as fast as Jill. What is the ratio of the kinetic energy of Jill to the kinetic energy of John?

Accepted Answer

Let's assume the mass of Jill to be 'm'. Therefore, the mass of John would be 0.5m.
Let's assume the speed of Jill to be 'v'. Therefore, the speed of John would be 2v.
Now, Kinetic Energy is given by 0.5*m*v^2
Let's calculate the KE of Jill and John and take their ratio:

KE of Jill = 0.5*m*v^2
KE of John = 0.5*0.5m*(2v)^2 = 2*0.5*m*v^2 = m*v^2

Ratio of KE of Jill to KE of John = KE of Jill/ KE of John = {0.5*m*v^2}/{m*v^2} = 0.5/1 = 1/2

Hence, the ratio of the Kinetic Energy of Jill to the Kinetic Energy of John is 1:2 or 1/2. Therefore, the answer is D.

Question 13

A 4.0 kg mass is moving with speed 2.0 m/s. A 1.0 kg mass is moving with speed 4.0 m/s. Both objects encounter the same constant braking force, and are brought to rest. Which object travels the greater distance before stopping?

Accepted Answer

Both objects experience the same braking force and the distance traveled before stopping depends only on their initial velocities, not on their masses. Therefore, both objects will travel the same distance before coming to a stop.

Question 14

You slam on the brakes of your car in a panic, and skid a certain distance on a straight, level road. If you had been traveling twice as fast, what distance would the car have skidded, under the same conditions?

Accepted Answer

The distance skidded by a car under braking on a level road is proportional to the square of the speed. Therefore, if the speed is doubled, the skidding distance becomes four times as far.

Question 15

A block of mass m is pushed against a spring of spring constant k. The spring is compressed by a distance d, the block is then released. It is launched by the spring along a horizontal frictionless surface with a final speed v. A second block, this one having mass 4m is pushed against the same spring by distance 6d and released. What is the final speed of the block in this case?

Accepted Answer

The potential energy stored in the spring is given by (1/2)kd^2. For the first block, this energy is converted into kinetic energy, (1/2)mv^2, giving v = sqrt(k/m) * d. For the second block, the spring is compressed by 6d, so the potential energy is (1/2)k(6d)^2 = 18kd^2. This energy is converted into kinetic energy of the second block, (1/2)(4m)v'^2. Solving for v', we get v' = sqrt(k/(4m)) * 6d = 3v.

Question 16

A car originally at rest has its speed increased to a value v in a period of 5 seconds. The work performed in this part of the motion is W₁. Over the next 5 seconds the speed of the car is increased to 2v. In the second part of the motion the work performed is W₂. It is correct to say that:

Accepted Answer

The answer of A car originally at rest has its...

Question 17

Consider a plot of the displacement (x) vs. applied force (F) for an ideal elastic spring. The slope of the curve would be

Accepted Answer

The answer of Consider a plot of the displacement (x)...

Question 18

A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much work in half as much time?

Accepted Answer

Using the formula for power:

P = W/T

Let x be the force that produces power P.

Then, the force that does six times as much work in half as much time would be 6x and T/2, respectively.

Using the formula for power again:

P' = (6W)/(T/2)

P' = (6W)(2/T)

P' = 12W/T

P' = 12P

Therefore, the power produced by the force that does six times as much work in half as much time is 12P, or choice letter A.

Question 19

FIGURE 7-1 
-Describe the type of spring "constant" needed to produce a constant restoring force like curve (a) in Figure 7-1.

Accepted Answer

To produce a constant restoring force, the spring constant k must vary inversely with the stretch, ensuring that the product of k and stretch remains constant.

Question 20

A block of mass m is pushed against a spring of spring constant k. The spring is compressed by a distance d, the block is then released. It is launched by the spring along a horizontal frictionless surface with a final speed v. A second block, this one having mass 4m is pushed against the same spring and released, gaining a final speed 2v. By what distance was the spring compressed in the second case?

Accepted Answer

Firstly, we can calculate the initial potential energy and final kinetic energy of the first block using the equation:
$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$
where x = d (distance compressed by the spring)

Then, we can find the spring constant for the second block by rearranging the equation above:
$k = \frac{2mv^2}{d^2}$

Next, we can use conservation of energy to find the distance compressed by the spring for the second block:
$\frac{1}{2}kx^2 = \frac{1}{2}(4m)(2v)^2$
$\frac{1}{2}\left(\frac{2mv^2}{d^2}\right)x^2 = 2mv^2$
$x^2 = 8d^2$
$x = \sqrt{8}d$
$x = 2\sqrt{2}d$

Therefore, the spring is compressed by 4d in the second case.

Quiz 7: Work and Kinetic Energy