Question 1

Two clowns are launched from the same spring-loaded circus cannon with the spring compressed the same distance each time. Clown A has a 40-kg mass; clown B a 60-kg mass. The relation between their kinetic energies at the instant of launch is:

Accepted Answer

The kinetic energy of an object is given by the equation KE = (1/2)mv^2, where m is the mass of the object and v is its velocity. Because the two clowns are launched with the spring compressed the same distance, they will have the same velocity when they leave the cannon. Therefore, the kinetic energy of each clown will be directly proportional to its mass. Since clown A has a mass of 40 kg and clown B has a mass of 60 kg, the relation between their kinetic energies at the instant of launch is KA/KB = (1/2)(40)/(1/2)(60) = 2/3. This means that KB is 1.5 times greater than KA. However, the question only asks for the relation between their kinetic energies, not their specific values. The only answer choice that accurately reflects this relationship is C, which states that KA = KB.

Question 2

Carts A and B have equal masses and travel equal distances D on side-by-side straight frictionless tracks while a constant force F acts on A and a constant force 2F acts on B. Both carts start from rest. The velocities $\overrightarrow{\mathbf{v}}$ A and $\overrightarrow{\mathbf{v}}$ B of the bodies at the end of distance D are related by:

Accepted Answer

The answer of Carts A and B have equal masses...

Question 3

The equation below is the solution to a physics problem: $\frac{1}{2}(2.30 \mathrm{~kg})\left(3.90 \frac{\mathrm{~m}}{\mathrm{~s}}\right)^{2}=\frac{1}{2}(2.30 \mathrm{~kg})\left(2.33 \frac{\mathrm{~m}}{\mathrm{~s}}\right)^{2}+(2.30 \mathrm{~kg})\left(9.80 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}\right)(1.00 \mathrm{~m}) \cos \left(60^{\circ}\right)$ .The most likely physical situation it describes is:

Accepted Answer

The equation given is the result of using Newton's second law (F=ma) along with the force of gravity (mg) and the force of friction (μN) on a cart of mass 2.30 kg rolling down a 30-degree incline. The negative sign on the right side of the equation indicates that the force is acting in the opposite direction of motion, which is downward. Therefore, the most likely physical situation is a 2.30 kg cart rolling down a 30-degree incline.

Question 4

Two balls, A and B, of mass m and 2m respectively, are carried to height h at constant velocity, but B rises twice as fast as A. The work the gravitational force does on B is:

Accepted Answer

The answer of Two balls, A and B, of mass...

Question 5

In a contest, two tractors pull two identical blocks of stone the same distance over identical surfaces. However, block A is moving twice as fast as block B when it crosses the finish line. Which statement is correct?

Accepted Answer

The answer of In a contest, two tractors pull two...

Question 6

Positive work can be done:

Accepted Answer

The answer of Positive work can be done:
...

Question 7

Planets go around the sun in elliptical orbits. The highly exaggerated diagram below shows a portion of such an orbit and the force on the planet at one position along that orbit. The planet is moving to the right. F|| and $F_{\perp}$ are the components of the force parallel (tangential) and perpendicular (normal) to the orbit. The work they do is W|| and $W_{\perp}$ . At the position shown:

Accepted Answer

The answer of Planets go around the sun in elliptical...

Question 8

The same constant force is used to accelerate two carts of the same mass, initially at rest, on horizontal frictionless tracks. The force is applied to cart A for twice as long a time as it is applied to cart B. The work the force does on A is WA; that on B is WB. Which statement is correct?

Accepted Answer

The answer of The same constant force is used to...

Question 9

Carts A and B have equal masses and travel equal distances on straight frictionless tracks while a constant force F is applied to A, and a constant force 2F is applied to B. The relative amounts of work done by the two forces are related by:

Accepted Answer

Work done is the product of force and distance. Since both carts travel equal distances, the work done on cart B (WB) is twice that on cart A (WA) because the force on B is 2F compared to F on A.

Question 10

Negative work can be done:

Accepted Answer

Negative work is done when the force applied is opposite to the direction of motion. In option (b), the spring applies an upward force while the elevator moves downward, and in option (c), the hand applies a force opposite to the ball's motion.

Question 11

Two identical springs with spring constant 50 N/m support a 5.0 N weight as in the picture below. What is the change in length of each spring when the weight is hung on the springs?

Accepted Answer

The answer of Two identical springs with spring constant 50...

Question 12

The graph below shows how the force on a 0.500 kg particle varies with position. If the particle has speed $v=2.23 \frac{\mathrm{~m}}{\mathrm{~s}}$ at x = 0.00 m, what is its speed in m/s when x = 8.00 m?

Accepted Answer

The answer of The graph below shows how the force...

Question 13

A 30-kg child sitting 5.0 m from the centre of a merry-go-round has a constant speed of 5.0 m/s. While she remains seated in the same spot and travels in a circle, the work the seat performs on her in one complete rotation is:

Accepted Answer

The work done by the seat on the child depends on the force applied and the displacement of the child. Since the child is not changing her position relative to the merry-go-round, the displacement is zero. Additionally, the force applied by the seat on the child is directed towards the centre of the merry-go-round, which is perpendicular to the displacement. Therefore, the work done by the seat on the child is zero.

Question 14

The force of static friction exerted on an automobile's tyres by the ground:

Accepted Answer

Static friction provides the force necessary for the car to accelerate forward, but it does not do work since there is no displacement at the point of contact.

Question 15

After a skydiver reaches terminal velocity:

Accepted Answer

When a skydiver reaches terminal velocity, the work done by the force of gravity is converted into thermal energy due to air resistance and friction, not into gravitational potential energy. This process increases the thermal energy of the skydiver and the surrounding air.

Question 16

Equal amounts of work are performed on two bodies, A and B, initially at rest, and of masses m and 2m respectively. The relation between their speeds immediately after the work has been done on them is:

Accepted Answer

The answer of Equal amounts of work are performed on...

Question 17

A group of rioting students throw rotten eggs at some passing politicians. Two eggs of equal mass are thrown with equal velocity. Egg B hits the soft padded shoulders of a politician's suit but egg A misses and hits the building wall behind them instead. Compare the work done on the eggs in reducing their velocities to zero.

Accepted Answer

The work done on both eggs is the same because they both come to a stop, meaning their kinetic energy is reduced to zero. The initial kinetic energy of both eggs is the same, so the work done to stop them is also the same.

Question 18

Two cannonballs are dropped from a second floor physics lab at height h above the ground. Ball B has four times the mass of ball A. When the balls pass the bottom of a first floor window at height $\frac{h}{4}$ above the ground, the relation between their kinetic energies, KA and KB, is:

Accepted Answer

The answer of Two cannonballs are dropped from a second...

Question 19

The work $\overrightarrow{\mathbf{F}}_{s p} \cdot d \overrightarrow{\mathbf{x}}$ done by the force exerted by the spring on a mass attached to the end of the spring when the mass has displacement d $\overrightarrow{\mathbf{x}}$ is

Accepted Answer

The answer of The work $\overrightarrow{\mathbf{F}}_{s p} \cdot d \overrightarrow{\mathbf{x}}$...

Question 20

Sally, who weighs 450 N, stands on a skate board while Roger pushes it forward 13.0 m at constant velocity on a level straight street. He applies a constant 100 N force.

Accepted Answer

Work is calculated as the product of the force applied in the direction of motion and the displacement. Since Roger applies a 100 N force over a distance of 13.0 m, the work done is $100 \, \text{N} \times 13.0 \, \text{m} = 1300 \, \text{J}$.

Quiz 6: Work, Force and Energy