Question 1

A 5.0-kg cart is moving horizontally at 6.0 m/s.In order to change its speed to 10.0 m/s, the net work done on the cart must be:

Accepted Answer

The net work done on an object is equal to the change in its kinetic energy. 
The initial kinetic energy of the cart is (1/2)mv^2 = (1/2)(5.0 kg)(6.0 m/s)^2 = 90 J.
The final kinetic energy of the cart is (1/2)mv^2 = (1/2)(5.0 kg)(10.0 m/s)^2 = 250 J.
Therefore, the change in kinetic energy is: 
250 J - 90 J = 160 J. 
Thus, the net work done on the cart must be 160 J, which is choice D.

Question 2

An 8-N block slides down an incline.It has an initial speed of 7 m/s.The work done by the resultant force on this block is:

Accepted Answer

The work done by the resultant force on the block cannot be determined with the information provided. Essential details such as the distance the block slides, the angle of the incline, and the coefficient of friction (if any) are missing. These factors are necessary to calculate the work done.

Question 3

The mechanical advantage of any machine is:

Accepted Answer

The mechanical advantage of a machine is defined as the ratio of the force exerted by the machine (output force) to the force applied to it (input force). It indicates how much the machine multiplies the force applied to it, making it easier to do work.

Question 4

A 4-kg cart starts up an incline with a speed of 3 m/s and comes to rest 2 m up the incline.The total work done on the cart is:

Accepted Answer

The total work done on the cart can be calculated using the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy. Initially, the cart has kinetic energy due to its speed, and it comes to rest at a higher elevation, which means all its initial kinetic energy is converted into potential energy (and possibly overcoming friction, but this is not specified). The initial kinetic energy (KE) of the cart is given by $KE = \frac{1}{2}mv^2$, where $m = 4 \, \text{kg}$ and $v = 3 \, \text{m/s}$. So, $KE = \frac{1}{2} \times 4 \times (3)^2 = 18 \, \text{J}$. Since the cart comes to rest, all this energy is used to move up the incline and possibly overcome friction. Therefore, the work done on the cart is negative (since it's losing kinetic energy), and the magnitude of this work is equal to the initial kinetic energy, which is -18 J.

Question 5

A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30°with the horizontal.His force is parallel to the slope.If the speed of the crate decreases at a rate of 1.5 m/s², then the work done by the man is:

Accepted Answer

The work done by the man is calculated using the formula: Work = Force x Distance. The force exerted by the man must overcome both the gravitational component along the slope and the deceleration of the crate. The gravitational force component along the slope is 80 N * sin(30°) = 40 N. The deceleration force is mass * deceleration = (80 N / 9.8 m/s²) * 1.5 m/s² = 12.24 N. The total force exerted by the man is 40 N + 12.24 N = 52.24 N. Therefore, the work done is 52.24 N * 5 m = 261.2 J. However, considering the options, the closest value is 140 J, which suggests a possible error in the problem setup or options.

Question 6

The amount of work required to stop a moving object is equal to the:

Accepted Answer

The amount of work required to stop a moving object is equal to its kinetic energy, which is represented by option B. This is because kinetic energy is the energy an object possesses due to its motion and represents the work that would be required to stop the object. The other options do not accurately represent the amount of work required to stop a moving object.

Question 7

Two objects with masses, m₁ and m₂, have the same kinetic energy and are both moving to the right.The same constant force is applied to the left to both masses.If m₁ = 4m₂, the ratio of the stopping distance of m₁ to that of m₂ is:

Accepted Answer

The answer of Two objects with masses, m₁ and m₂,...

Question 8

The velocity of a particle moving along the x axis changes from v_i to v_f.For which values of v_i and v_f is the total work done on the particle positive?

Accepted Answer

The answer of The velocity of a particle moving along...

Question 9

An 80-N crate slides with constant speed a distance of 5.0 m downward along a rough slope that makes an angle of 30° with the horizontal.The work done by the force of gravity is:

Accepted Answer

The work done by the force of gravity can be calculated using the formula $W = Fd\cos(\theta)$, where $F$ is the force of gravity (weight of the crate), $d$ is the distance moved, and $\theta$ is the angle between the force and the direction of motion. Since the crate is moving downward along the slope and gravity acts vertically downward, the angle to consider here is the slope's angle with the horizontal, which is 30 degrees. However, the effective component of gravity that does the work is along the slope, so we use the component of the gravitational force parallel to the slope. This component is $F\cos(30^\circ)$, but since the motion is directly down the slope, we actually use the full force of gravity for calculating work done along the slope, which simplifies to $W = Fd$, because the angle between the direction of gravity's force (downward) and the direction of motion (also downward along the slope) effectively becomes 0 degrees in terms of the component of force doing work, meaning $\cos(0^\circ) = 1$. Therefore, the calculation should directly use the gravitational force and the distance moved without needing to adjust for the angle in this context. The gravitational force is the weight of the crate, which is given as 80 N, and the distance is 5.0 m. Thus, the work done by gravity is $W = 80\, \text{N} \times 5.0\, \text{m} = 400\, \text{J}$. The work is positive because the force of gravity and the direction of motion are in the same direction, indicating that gravity is doing work on the object to move it downward.

Question 10

At time t = 0 a 2-kg particle has a velocity of (4 m/s)  - (3 m/s)  .At t = 3 s its velocity is (2 m/s)  + (3 m/s)  .During this time the work done on it was:

Accepted Answer

The answer of At time t = 0 a 2-kg...

Question 11

A particle starts from rest at time t = 0 and moves along the x axis.If the net force on it is proportional to t, its kinetic energy is proportional to:

Accepted Answer

When the net force on a particle is proportional to t, we can write the force as F = kt, where k is a constant. By Newton's Second Law, we know that F = ma, or kt = ma. Rearranging, we get a = (k/m)t.

We can integrate this acceleration function to find the particle's velocity as a function of time:

v(t) = a∫dt = (k/m)∫t dt = (k/m)(t^2/2) + C

Using the initial condition that the particle is at rest at t = 0, we can solve for C:

v(0) = 0 = (k/m)(0^2/2) + C 
C = 0

So the particle's velocity as a function of time is:

v(t) = (k/m)(t^2/2)

Now we can find the kinetic energy of the particle as a function of time:

KE(t) = (1/2)mv(t)^2 
KE(t) = (1/2)mk^2(t^4/4m^2) 
KE(t) = (k^2/8m)(t^4)

Since k and m are constants, we see that the kinetic energy is proportional to t^4. Therefore, the answer is (C) t^(1/4).

Question 12

In raising an object to a given height by means of an inclined plane, as compared with raising the object vertically, there is a reduction in:

Accepted Answer

Using an inclined plane reduces the force required to lift an object to a certain height, but does not change the work done, distance (which actually increases), friction inherently, or the value of acceleration due to gravity.

Question 13

A 1-kg block is lifted vertically 1 m at constant speed by a boy.The work done by the boy is about:

Accepted Answer

The work done by the boy is given by the formula:
Work = Force x Distance x cos(theta)
However, since the block is lifted at constant speed, the force applied by the boy is equal and opposite to the force of gravity on the block. Therefore, the net work done on the block is zero. 
However, the work done by the boy is still equal to the work done against gravity, which is given by:
Work = mgh
where m = 1 kg, g = 9.8 m/s^2 (acceleration due to gravity), and h = 1 m
Plugging in these values, we get:
Work = (1 kg) x (9.8 m/s^2) x (1 m) = 9.8 J, which is approximately equal to 10 J (choice C).

Question 14

A 100-kg piano rolls down a 20° incline.A man tries to keep it from accelerating, and manages to keep its acceleration to 1.2 m/s².If the piano rolls 5 m, what is the net work done on it by all the forces acting on it?

Accepted Answer

First, we need to find the total force acting on the piano while it rolls down the incline. We can use the formula:

force = mass x acceleration

The mass of the piano is 100 kg, and the acceleration is given as 1.2 m/s^2. Therefore, the force is:

force = 100 kg x 1.2 m/s^2 = 120 N

Next, we need to find the work done by this force over a distance of 5 m. We can use the formula:

work = force x distance x cos(theta)

where theta is the angle between the force and the displacement. Since the force and the displacement are in the same direction, we can simplify this to:

work = force x distance

Plugging in the values, we get:

work = 120 N x 5 m = 600 J

Therefore, the net work done on the piano by all the forces acting on it is 600 J, which corresponds to choice letter C.

Question 15

A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m.An external force of 3.0 N, always tangent to the track, causes the object to speed up as it goes around.If it starts from rest its speed at the end of one revolution is:

Accepted Answer

The answer of A 0.50-kg object moves in a horizontal...

Question 16

A Boston Red Sox baseball player catches a ball of mass m that is moving toward him with speed v.While bringing the ball to rest, his hand moves back a distance d.Assuming constant deceleration, the horizontal force exerted on the ball by his hand is:

Accepted Answer

The work done by the player's hand in bringing the ball to rest is given by the work-energy theorem: W = ΔK, where W is the work done by the force exerted by the hand, ΔK is the change in the kinetic energy of the ball, and we assume that all the work done by the hand is transferred to the ball. The initial kinetic energy of the ball is K1 = (1/2)mv^2, and the final kinetic energy is K2 = 0. The distance over which the force is applied is d, and the deceleration is a. Therefore, we have: W = ΔK = K2 - K1 = -(1/2)mv^2 The work done by the hand is also given by the formula: W = Fd where F is the force exerted by the hand, and d is the distance over which it is applied. Therefore, we have: Fd = -(1/2)mv^2 Solving for F, we get: F = -(1/2)mv^2/d = -mv^2/(2d) Note that the negative sign just indicates that the force is opposite to the direction of motion of the ball. To find the magnitude of the force, we take the absolute value: |F| = mv^2/(2d) Therefore, the horizontal force exerted on the ball by the player's hand is given by: Answer: E) mv²/(2d)

Question 17

The work done by gravity during the descent of a projectile is:

Accepted Answer

The work done by gravity on a descending projectile is positive because the force of gravity and the direction of motion are in the same direction, resulting in positive work.

Question 18

A 0.50-kg object moves on a horizontal frictionless circular track with a radius of 2.5 m.An external force of 3.0 N, always tangent to the track, causes the object to speed up as it goes around.If it starts from rest, then at the end of one revolution the radial component of the force of the track on it is:

Accepted Answer

The radial force is the centripetal force required to keep the object moving in a circle. After one revolution, the work done by the tangential force equals the change in kinetic energy. Using work-energy principle and centripetal force formula, the radial force is calculated as 38 N.

Question 19

A man pulls a 100-N crate up a frictionless 30°slope 5 m high as shown.Assuming that the crate moves at constant speed, the work done by the man is:

Accepted Answer

The answer of A man pulls a 100-N crate up...

Question 20

At time t = 0 a particle starts moving along the x axis.If its kinetic energy increases uniformly with t the net force acting on it must be:

Accepted Answer

The answer of At time t = 0 a particle...

Quiz 7: Kinetic Energy and Work