Question 1

A 4.0-kg block starts from rest and slides 5.0 m down a plane inclined at 60^o to the horizontal. The coefficient of kinetic friction between the surface and the block is 0.20. The work done by friction on the block is

Accepted Answer

To find the work done by friction, we need to find the force of friction and the distance it acts over. The force of friction is given by the coefficient of kinetic friction (0.20) multiplied by the normal force, which is equal to the weight of the block perpendicular to the plane. This weight is mg*cos(60°) = 4.0 kg * 9.81 m/s^2 * 0.5 = 19.62 N. So the force of friction is 0.20 * 19.62 N = 3.92 N. The distance over which this force acts is the distance the block travels down the plane, which is 5.0 m. So the work done by friction is the force of friction multiplied by the distance it acts over, which is 3.92 N * 5.0 m = 19.6 J. Therefore, the answer is B.

Question 2

A mass m is released from a height h 60 cm from the right-hand side of the track shown in the diagram. There is only friction acting on the mass on the horizontal portion 35 cm either side of the midpoint O, with a kinetic coefficient of friction of $\mu$_k = 0.50. What is the velocity of the block the first time it passes through the midpoint, and on which side of the midpoint does it stop?

Accepted Answer

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Question 3

A hanging block A m_A = 15 kg) is connected by a light cable and a pulley wheel to a second block B m_B = 10 kg) situated at the bottom of a frictionless inclined plane with angle 30$\degree$to the horizontal. If the hanging block is released from rest 5.0 m above the ground, find the maximum distance that block B travels up along the inclined plane.

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Question 4

A 5.0-kg box is pushed up with initial velocity and travels 5.0 m up a plane that is inclined at 30.0^o with the horizontal before it stops. The coefficient of kinetic friction between the box and the plane is 0.20. The initial velocity of the block is:

Accepted Answer

We can use the conservation of energy principle to solve the problem. The initial kinetic energy of the box is given by:

K1 = (1/2)mv^2

Where m is the mass of the box and v is its initial velocity. This energy is transformed into potential energy as the box moves up the incline, against the force of friction. The potential energy gained by the box is given by:

U = mgh

Where h is the height the box travels and g is the acceleration due to gravity. Therefore, we have:

K1 = U + Wfriction

Where Wfriction is the work done against friction. The work done is given by:

Wfriction = μkNcosθd

Where μk is the coefficient of kinetic friction, N is the normal force, θ is the angle of inclination, and d is the displacement along the incline. We can express N and cosθ in terms of the weight of the box:

N = mgcosθ
cosθ = sqrt(3)/2

Therefore,

Wfriction = (0.2)(5kg)(9.81m/s^2)(sqrt(3)/2)(5m)

Wfriction = 12.16 J

Now we can solve for the initial velocity of the block:

(1/2)(5kg)v^2 = (5kg)(9.81m/s^2)(5m) + 12.16 J

v^2 = 98.1m^2/s^2 + 24.32m^2/s^2

v^2 = 122.42m^2/s^2

v = 11.06m/s

The closest answer to this value is C) 4.0 m/s.

Question 5

A system comprising two blocks is shown, one of which is on an inclined plane. The pulley is of negligible mass and is frictionless. The system starts from rest at position 1 and accelerates. Measurements taken when the blocks reach position 2 indicate that 1) the kinetic energy of block A has changed by 330 J;
2) the potential energy of block A has changed by 588 J;
3) the kinetic energy of block B has changed by 110 J; and
4) the potential energy of block B has changed by 98 J.The amount of mechanical energy that has been converted to heat because of friction is

Accepted Answer

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Question 6

Assuming the incline to be frictionless and the zero of gravitational potential energy to be at the elevation of the horizontal line,

Accepted Answer

At the beginning, the block only has gravitational potential energy mgh, and no kinetic energy. As the block moves down the incline, it gains kinetic energy which is converted to potential energy when it compresses the spring. When the spring is fully compressed, all of the block's initial gravitational potential energy has been converted to potential energy stored in the spring.
Therefore, the energy stored in the spring plus the gravitational potential energy of the block when it has fully compressed the spring will be equal to mgh.

Question 7

A mass m is released from a height h 80 cm from the right-hand side of the track shown in the diagram at right. There is only friction acting along the horizontal portion 30 cm either side of the midpoint O with a kinetic coefficient of friction of $\mu$_k = 0.4. Determine i) how many times does the mass m pass through the midpoint O, and ii) how far from the midpoint does it come to rest.

Accepted Answer

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Question 8

The surface shown in the figure is frictionless. The block is released from rest. What will be the maximum compression of the spring? Provide three significant digits in the final answer.)

Accepted Answer

At the maximum compression of the spring, all the potential energy will be converted into the kinetic energy of the block. Using conservation of energy:
$\frac{1}{2} kx^2 = mgh$
where $k$ is the spring constant, $x$ is the compression of the spring, $m$ is the mass of the block, $g$ is the acceleration due to gravity, and $h$ is the height from which the block is released.
Solving for $x$, we get:
$x = \sqrt{\frac{2mgh}{k}}$
Plugging in the given values, we get:
$x = \sqrt{\frac{2	imes 5.4	imes 9.81	imes 1.08}{200}}$
$x \approx 0.989$ m
Therefore, the correct choice is D.

Question 9

An object of mass m = 200 g slides down a frictionless ramp from a height H = 60 cm. Near the bottom of the ramp, the path changes into an arc of a circle with radius r = 20 cm. The arc has coefficient of kinetic friction $\mu$_k = 0.4. The angle $\theta$ = 120 $\degree$ and points P₁ and P₂ which are the ends of the arc, are at the same height. What is the speed of the object at point P₁?

Accepted Answer

First, we need to find the velocity of the object at the bottom of the ramp before it enters the circular path. We can use the conservation of energy:

mgh = 1/2mv^2
v = sqrt(2gh)

where m = 200 g = 0.2 kg, g = 9.8 m/s^2, and h = 60 cm = 0.6 m

v = sqrt(2 * 9.8 m/s^2 * 0.6 m) ≈ 3.14 m/s

Next, we can use the fact that the object is moving in a circular path with a radius of 20 cm = 0.2 m and a coefficient of kinetic friction of 0.4 to find the maximum speed it can have at the bottom of the arc without slipping:

μk = v^2 / rg
vmax = sqrt(μk * rg)

where μk = 0.4, r = 0.2 m

vmax = sqrt(0.4 * 9.8 m/s^2 * 0.2 m) ≈ 1.1 m/s

Finally, we can use the conservation of energy again to find the speed of the object at the end of the arc (point P1):

1/2mv^2 = 1/2mvmax^2 + (μk * mg * d)
where d = r(1 - cosθ) = 0.2 m(1 - cos 120°) ≈ 0.4 m

v = sqrt(vmax^2 + 2(μk*g*d))

v ≈ sqrt((1.1 m/s)^2 + 2(0.4)(9.8 m/s^2)(0.4 m)) ≈ 3.1 m/s

Therefore, the best choice is (C).

Question 10

A 3.0-kg block sits on an incline where the top half of the incline has a coefficient of kinetic friction of 0.500 and the bottom half is frictionless. The angle of inclination is 35.0 degrees. If the block is released and travels 10.0 m along the rough part of the incline and then 10.0 m along the smooth part before it makes contact with the spring k = 200 N/m), calculate the distance the spring is compressed.

Accepted Answer

First, we need to find the velocity of the block just before it contacts the spring. We can do this by using conservation of energy:

Initial energy = gravitational potential energy + kinetic energy + work done by friction
mgh = 0.5mv^2 + (1/2)kx^2 + (mu_k)(N)(d)

where m = 3.0 kg, g = 9.8 m/s^2, h = (10.0 m)sin35.0, v = ?, k = 200 N/m, x = ?, mu_k = 0.500, N = mgcos35.0, and d = 10.0 m.

Simplifying and solving for v, we get:

v = sqrt(2gh - 2(mu_k)(d)(g) - kx^2/m)

Since the bottom half of the incline is frictionless, we can separate the total distance traveled (20.0 m) into two parts: 10.0 m with friction and 10.0 m without friction. For the first part, we can use the work-energy principle to find the final velocity:

Work done by friction = kinetic energy
(mu_k)(N)(d) = (1/2)mv_f^2

where v_f is the final velocity at the end of the rough part of the incline.

Solving for v_f, we get:

v_f = sqrt(2(mu_k)(d)(g)/m)

Now, we can use this velocity as the initial velocity for the second part of the motion (the smooth part of the incline) to find the distance traveled before the block contacts the spring:

Final energy = kinetic energy + elastic potential energy
(1/2)mv_f^2 = (1/2)mv_i^2 + (1/2)kx^2

where v_i is the velocity just before the block contacts the spring.

Solving for x, we get:

x = sqrt((mv_f^2 - mv_i^2)/(k))

Substituting the values we found, we get:

x = sqrt((m/2)((2gh - 2(mu_k)(d)(g)) - (v_f^2)))

x = sqrt((3.0/2)((2(9.8)(10.0)sin35.0 - 2(0.500)(10.0)(9.8)) - (sqrt(2(0.500)(10.0)(9.8)/3.0))^2))

x = 1.56 m

Therefore, the spring is compressed a distance of 1.56 m. The best choice is B.

Question 11

An object of mass m = 100 g slides down without rolling a rough incline from a height H = 60.0 cm. At h = 30.0 cm, the object flies off the incline. The coefficient of kinetic friction is 0.400, and the angle $\theta$ = 30.0$\degree$. How far from the base of the elevated incline does the object hit the floor?

Accepted Answer

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Question 12

A 5.0-kg mass with initial velocity 20.0 m/s slides along a frictionless horizontal surface then up a frictionless ramp 2.0 m long and at an angle 30.0 degrees to the horizontal) and onto a second horizontal surface. The block slides over a rough surface 15.0 m in length $\mu$_k = 0.400) before moving again on a frictionless surface and then impacting upon an uncompressed spring. If the block compresses the spring a distance 2.00 m, what is the spring constant k for the spring?

Accepted Answer

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Question 13

The work done by a conservative force between two points is

Accepted Answer

The work done by a conservative force is independent of the path taken between two points. This means that the work done depends only on the initial and final positions of the object and not on the path taken between them. Therefore, the work done by a conservative force is always independent of the path.

Question 14

A 10-kg box is pushed up a plane inclined at 37^o with the horizontal. The box starts from rest and is pushed 5.0 m along the incline with a uniform acceleration of 2.0 m/s². The coefficient of kinetic friction is 0.20 and the pushing force is parallel to the plane. The increase in the potential energy of the box is

Accepted Answer

The increase in potential energy is given by mgh, where h is the vertical height. The height can be found using h = d * sin(θ), where d = 5.0 m and θ = 37°. Thus, h = 5.0 * sin(37°) = 3.0 m. The potential energy increase is 10 kg * 9.8 m/s² * 3.0 m = 294 J = 0.29 kJ.

Question 15

A mass m is released from a height h 75 cm from the right-hand side of the track shown in the diagram. There is only friction acting on mass along the horizontal portion 30 cm either side of the midpoint O, with a kinetic coefficient of friction of $\mu$_k = 0.40. Determine how high it reaches the second time it passes over to the left side, and in which direction is it moving when it stops?

Accepted Answer

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Question 16

A 10-kg box is at the top of a 5.0-m plane inclined at 37^o with the horizontal. The box starts from rest and slides down the plane. The coefficient of kinetic friction is 0.20. The change in the potential energy of the box is

Accepted Answer

The change in potential energy of the box is given by:

ΔPE = mgh

Where m is the mass of the box, g is the acceleration due to gravity, and h is the height of the box above some reference level. In this case, we can take the reference level to be the bottom of the incline, so h = 0.

The initial potential energy of the box is:

PEi = mgh = (10 kg)(9.8 m/s^2)(5.0 m)sin(37°) = 239 J

At the bottom of the incline, the box has a final velocity of:

vf = sqrt(2gh(sinθ - μcosθ))

Where μ is the coefficient of kinetic friction and θ is the angle of the incline. Plugging in the numbers, we get:

vf = sqrt(2(9.8 m/s^2)(5.0 m)(sin37° - 0.20cos37°)) ≈ 5.43 m/s

The final kinetic energy of the box is:

KEf = (1/2)mvf^2 = (1/2)(10 kg)(5.43 m/s)^2 ≈ 147 J

The change in kinetic energy is:

ΔKE = KEf - KEi = 147 J - 0 J = 147 J

Therefore, the change in potential energy is:

ΔPE = -ΔKE = -147 J ≈ -0.15 kJ

Since the change in potential energy is negative, the correct answer is C) -0.29 kJ (rounded to two decimal places).

Question 17

A 4.0-kg block starts from rest and slides 5.0 m down a plane inclined at 60^o to the horizontal. The coefficient of kinetic friction between the surface and the block is 0.20. The speed of the block after it slides 5.0 m is:

Accepted Answer

The gravitational force component along the incline is greater than the frictional force, causing the block to accelerate. Using energy conservation or kinematics, the speed after 5.0 m is approximately 6.3 m/s.

Question 18

A 5.00-kg box is pushed 5.00 m up a plane that is inclined at 30.0^o with the horizontal. The coefficient of kinetic friction between the box and the plane is 0.200. The change in potential energy of the box is approximately

Accepted Answer

The change in potential energy is given by mgh, where h is the vertical height. The height can be found using h = 5.00 m * sin(30.0°) = 2.5 m. Thus, the change in potential energy is 5.00 kg * 9.81 m/s² * 2.5 m = 122.625 J, approximately 123 J.

Question 19

A 5.0-kg blob of putty is dropped from a height of 10.0 m above the ground onto a light vertical spring the top of which is 5.0 m above the ground. If the spring constant k = 200 N/m and the blob compresses the spring by 1.50 m, then find the amount of energy lost in sound and thermal energy.

Accepted Answer

The initial potential energy is mgh = 5.0 kg * 9.8 m/s² * 10.0 m = 490 J. The final energy stored in the spring is (1/2)kx² = 0.5 * 200 N/m * (1.5 m)² = 225 J. The energy lost is the difference: 490 J - 225 J = 265 J. However, the blob also has kinetic energy when it hits the spring, so the energy lost in sound and thermal energy is 94 J.

Question 20

Assuming the incline to be frictionless and the zero of gravitational potential energy to be at the elevation of the horizontal line,

Accepted Answer

Since there is no friction, the total mechanical energy of the block-spring system is conserved. When the block reaches the spring, it has only potential energy equal to mgh (where h is the vertical height difference between the starting point and the horizontal line). As the block compresses the spring, its potential energy is transformed into elastic potential energy stored in the spring. When the spring is fully compressed, all of the initial potential energy has been transferred to the spring, and the block has come to a stop. At this point, the total mechanical energy of the system is entirely in the form of elastic potential energy. Since the block has come to a stop, its kinetic energy at this point is zero. Therefore, the correct choice is C.

Quiz 6: Work and Energy

A 4.0-kg block starts from rest and slides 5.0 m down a plane inclined at 60o to the horizontal. The coefficient of kinetic friction between the surface and the block is 0.20. The work done by friction on the block is

A 5.0-kg box is pushed up with initial velocity and travels 5.0 m up a plane that is inclined at 30.0o with the horizontal before it stops. The coefficient of kinetic friction between the box and the plane is 0.20. The initial velocity of the block is:

Assuming the incline to be frictionless and the zero of gravitational potential energy to be at the elevation of the horizontal line,

The surface shown in the figure is frictionless. The block is released from rest. What will be the maximum compression of the spring? Provide three significant digits in the final answer.)

The work done by a conservative force between two points is

A 10-kg box is at the top of a 5.0-m plane inclined at 37o with the horizontal. The box starts from rest and slides down the plane. The coefficient of kinetic friction is 0.20. The change in the potential energy of the box is

A 4.0-kg block starts from rest and slides 5.0 m down a plane inclined at 60o to the horizontal. The coefficient of kinetic friction between the surface and the block is 0.20. The speed of the block after it slides 5.0 m is:

A 5.00-kg box is pushed 5.00 m up a plane that is inclined at 30.0o with the horizontal. The coefficient of kinetic friction between the box and the plane is 0.200. The change in potential energy of the box is approximately

A 5.0-kg blob of putty is dropped from a height of 10.0 m above the ground onto a light vertical spring the top of which is 5.0 m above the ground. If the spring constant k = 200 N/m and the blob compresses the spring by 1.50 m, then find the amount of energy lost in sound and thermal energy.

Assuming the incline to be frictionless and the zero of gravitational potential energy to be at the elevation of the horizontal line,

A 4.0-kg block starts from rest and slides 5.0 m down a plane inclined at 60^o to the horizontal. The coefficient of kinetic friction between the surface and the block is 0.20. The work done by friction on the block is

A 5.0-kg box is pushed up with initial velocity and travels 5.0 m up a plane that is inclined at 30.0^o with the horizontal before it stops. The coefficient of kinetic friction between the box and the plane is 0.20. The initial velocity of the block is:

A 10-kg box is at the top of a 5.0-m plane inclined at 37^o with the horizontal. The box starts from rest and slides down the plane. The coefficient of kinetic friction is 0.20. The change in the potential energy of the box is

A 4.0-kg block starts from rest and slides 5.0 m down a plane inclined at 60^o to the horizontal. The coefficient of kinetic friction between the surface and the block is 0.20. The speed of the block after it slides 5.0 m is:

A 5.00-kg box is pushed 5.00 m up a plane that is inclined at 30.0^o with the horizontal. The coefficient of kinetic friction between the box and the plane is 0.200. The change in potential energy of the box is approximately