Question 1

Objects A and B,of mass M and 2M respectively,are each pushed a distance d straight up an inclined plane by a force F parallel to the plane.The coefficient of kinetic friction between each mass and the plane has the same value $\mu$_k.At the highest point,

Accepted Answer

The answer of Objects A and B,of mass M and...

Question 2

An all-terrain vehicle of 2 000 kg mass moves up a 15.0$\degree$ slope at a constant velocity of 6.00 m/s.The rate of change of gravitational potential energy with time is

Accepted Answer

The answer of An all-terrain vehicle of 2 000 kg...

Question 3

A champion athlete can produce one horsepower (746 W)for a short period of time.If a 70-kg athlete were to bicycle to the summit of a 500-m high mountain while expending power at this rate,she would have used at least ____ J of energy.

Accepted Answer

The answer of A champion athlete can produce one horsepower...

Question 4

As a result of friction between internal parts of an isolated system

Accepted Answer

Friction always causes a loss of mechanical energy, meaning the total mechanical energy of the system decreases. This loss of energy is often dissipated as heat, sound or other forms of energy that are not useful for the system's motion.

Question 5

A 2.0-kg mass is projected vertically upward from ground level with an initial speed of 30 m/s.The mass rises to a maximum height of 35 m above ground level.What is the change in mechanical energy of the mass caused by air resistance between the point of projection and the point of maximum height?

Accepted Answer

Explanation:The change in mechanical energy is equal to the work done by air resistance, which is given by:$$W = Fd = mgd$$where:* F is the force of air resistance* d is the distance over which the force acts* m is the mass of the object* g is the acceleration due to gravityIn this case, the distance over which the force acts is equal to the maximum height reached by the object, which is 35 m. The force of air resistance is equal to the weight of the object, which is:$$F = mg = (2.0 	ext{ kg})(9.8 	ext{ m/s}^2) = 19.6 	ext{ N}$$Therefore, the change in mechanical energy is:$$W = Fd = (19.6 	ext{ N})(35 	ext{ m}) = 686 	ext{ J}$$Converting to kilojoules, we get:$$W = 686 	ext{ J} = 0.686 	ext{ kJ}$$Therefore, the correct answer is A) 0.686 kJ.

Question 6

A 6.0-kg block slides along a horizontal surface.If µ_k = 0.20 for the block and surface,at what rate is the friction force changing the mechanical energy of the block at an instant when its speed is 4.0 m/s?

Accepted Answer

The answer of A 6.0-kg block slides along a horizontal...

Question 7

A spring with spring constant 800 N/m compressed 0.200 m is released and projects a 0.800 kg mass along a frictionless surface.The mass reaches a surface area where $\mu$_k = 0.400 and comes to a stop.The following student solution contains at least one error.What is the error? $$\frac { 1 } { 2 } \left( 800 \frac { \mathrm { N } } { \mathrm { m } } \right) ( 0.200 \mathrm {~m} ) ^ { 2 } = \frac { 1 } { 2 } ( 0.500 \mathrm {~kg} ) \left( 8.00 \frac { \mathrm { m } } { \mathrm { s } } \right) ^ { 2 } + 0.400 ( 0.500 \mathrm {~kg} ) \left( 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } } \right) ( 8.16 \mathrm {~m} )$$

Accepted Answer

The elastic potential energy is converted into kinetic energy and then into internal thermal energy due to friction, but not both simultaneously.

Question 8

As an object moves from point A to point B only two forces act on it: one force is conservative and does -70 J of work,the other force is nonconservative and does +50 J of work.Between A and B,

Accepted Answer

The answer of As an object moves from point A...

Question 9

A 25-kg block on a horizontal surface is attached to a light spring (force constant = 8.0 kN/m).The block is pulled 10 cm to the right from its equilibrium position and released from rest.When the block has moved 2.0 cm toward its equilibrium position,its kinetic energy is 12 J.What is the change in mechanical energy caused by the frictional force on the block as it moves the 2.0 cm?

Accepted Answer

The change in mechanical energy is equal to the work done by all forces except the spring force. We can find the work done by the frictional force by using the work-energy theorem: 
Work done by frictional force = kinetic energy at 2.0 cm - kinetic energy at 10 cm 
= 12 J - (1/2)(25 kg)((0.10 m)^2)((8.0 kN/m)/(1000 N/kN)) 
= 12 J - 1 J 
= 11 J

Therefore, the change in mechanical energy caused by the frictional force on the block is 11 J. Choice C is the correct answer.

Question 10

A 2.0-kg block is projected down a plane that makes an angle of 20$\degree$ with the horizontal with an initial kinetic energy of 2.0 J.If the coefficient of kinetic friction between the block and plane is 0.40,how far will the block slide down the plane before coming to rest?

Accepted Answer

Explanation:The work done by friction is:$$W_f=f_kd=\mu_kmgd\sin	heta$$The initial kinetic energy is:$$K_i=\frac{1}{2}mv_i^2$$The final kinetic energy is:$$K_f=0$$The work-energy theorem states that:$$W_{net}=K_f-K_i$$Substituting in the expressions for the work done by friction and the initial and final kinetic energies, we get:$$\mu_kmgd\sin	heta=\frac{1}{2}mv_i^2$$Solving for d, we get:$$d=\frac{v_i^2}{2g\mu_k\sin	heta}$$Substituting in the given values, we get:$$d=\frac{(2.0	ext{ m/s})^2}{2(9.8	ext{ m/s}^2)(0.40)\sin20^\circ}=3.0	ext{ m}$$

Question 11

A champion athlete can produce one horsepower (746 W)for a short period of time.If a 70-kg athlete were to bicycle to the summit of a 500-m high mountain while expending power at this rate,she would reach the summit in ____ seconds.

Accepted Answer

The work done by the athlete is given by:
Work = force x distance
Here, force = power / velocity (since work = force x distance, and velocity = distance/time, therefore force = work/velocity)
Therefore, force = (746 W) / v
Also, since power = force x velocity,
velocity = power / force
Substituting the value of force in terms of velocity, we get:
velocity = power / (746 W / v) = (746 W x v) / 746 W = v
This means that the velocity of the athlete is constant.
Now, the work done by the athlete to climb the mountain is given by:
Work = force x distance
= (746 W / v) x 500 m
= (746 W / v) x (500 / v) m^2/s^2
= (746/500) J, since v is in m/s and the 746 W and 500 m cancel out.

Since the athlete's weight is 70 kg, the gravitational force on her is given by:
force = mass x acceleration due to gravity
= 70 kg x 9.81 m/s^2
= 686.7 N

The athlete's power output is given by:
Power = work / time
Therefore, time = work / power
Substituting the values, we get:
time = [(746/500) J] / (746 W)
= 1.492 s

Therefore, the time taken by the athlete to climb the mountain is 1.492 seconds, which is closest to option B (460 seconds).

Question 12

As an object moves from point A to point B only two forces act on it: one force is nonconservative and does -30 J of work,the other force is conservative and does +50 J of work.Between A and B,

Accepted Answer

The nonconservative force does negative work, which means it takes energy away from the object. The conservative force does positive work, which means it adds energy to the object. The total work done on the object is F1S1F1030 J + 50 J = -280 J. This means that the mechanical energy of the object decreases by 280 J. The kinetic energy of the object must also decrease, since the mechanical energy is the sum of the kinetic and potential energies.

Question 13

A 3.50 kg block is pulled along a moving conveyor belt at a constant speed of 0.500 m/s relative to a stationary observer while the belt moves at a constant speed of 0.200 m/s in the same direction.If the coefficient of kinetic friction is 0.400,the magnitude of the mechanical energy dissipated,in J,caused by the force of friction on the block in 8.00 s is

Accepted Answer

First, we need to find the force of friction acting on the block. The formula for frictional force is:

$f_k = \mu_k N$

where $f_k$ is the kinetic frictional force, $\mu_k$ is the coefficient of kinetic friction, and $N$ is the normal force.

The normal force $N$ is equal to the weight of the block since it is on a horizontal surface. Thus,

$N = mg = (3.50 kg)(9.81 m/s^2) = 34.335 N$

And so,

$f_k = (0.400)(34.335 N) = 13.734 N$

The work done by friction over a distance $d$ is given by:

$W_f = f_k d$

Here, we can use the fact that the block moves at a constant speed relative to the observer. We can assume that the block stays in the same position relative to the conveyor belt (i.e. it is not slipping) and therefore moves with the belt. Thus, over a time interval of $t = 8.00 s$, the block travels a distance of:

$d = v_{b,rel} t = (0.500 m/s - 0.200 m/s)(8.00 s) = 2.40 m$

And so, the work done by friction is:

$W_f = (13.734 N)(2.40 m) = 32.9616 J$

Therefore, the magnitude of the mechanical energy dissipated by friction in 8.00 s is 32.9 J, and the best choice is (C).

Question 14

The equation below describes a physical situation: $$\frac { 1 } { 2 } ( 1.70 \mathrm {~kg} ) \left( 3.30 \frac { \mathrm { m } } { \mathrm { s } } \right) ^ { 2 } + ( 1.70 \mathrm {~kg} ) \left( 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } } \right) ( 2.35 \mathrm {~m} ) \sin 30 ^ { \circ }$$ $$= \frac { 1 } { 2 } ( 1.70 \mathrm {~kg} ) \left( 4.60 \frac { \mathrm { m } } { \mathrm { s } } \right) ^ { 2 } + 0.320 ( 1.70 \mathrm {~kg} ) \left( 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } } \right) ( 2.35 \mathrm {~m} ) \cos 30 ^ { \circ }$$ Which description best fits the equation?

Accepted Answer

The answer of The equation below describes a physical situation:...

Question 15

The two masses in the figure are released from rest.After the 3.0-kg mass has fallen 1.5 m,it is moving with a speed of 3.8 m/s.What is the change in mechanical energy done on the system during this time interval by the frictional force on the 2.0 kg mass?

Accepted Answer

The answer of The two masses in the figure are...

Question 16

A 3.50 kg block is pulled along a moving conveyor belt at a constant speed of 0.500 m/s relative to a stationary observer while the belt moves at a constant speed of 0.200 m/s in the opposite direction.If the coefficient of kinetic friction is 0.400,the magnitude of the mechanical energy dissipated,in J,caused by the force of friction on the block in 8.00 s is

Accepted Answer

The mechanical energy dissipated due to friction is calculated using the work done by the friction force over a distance. The friction force can be found using the formula $f = \mu_k N$, where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force. Since the block is moving at a constant speed, the normal force equals the weight of the block, $N = mg$, where $m = 3.50 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$.First, calculate the friction force:$$f = \mu_k mg = 0.400 \times 3.50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 13.72 \, \text{N}$$The distance the block moves relative to the conveyor belt in 8.00 s is the relative speed multiplied by time. However, since the question asks for the energy dissipated due to friction, we need to consider the distance the block moves relative to a stationary observer. The speed of the block relative to the stationary observer is the sum of the speeds of the block and the conveyor belt (since they move in opposite directions), which is $0.500 \, \text{m/s} + 0.200 \, \text{m/s} = 0.700 \, \text{m/s}$.The distance moved in 8.00 s is:$$d = vt = 0.700 \, \text{m/s} \times 8.00 \, \text{s} = 5.6 \, \text{m}$$The work done by the friction force, which equals the mechanical energy dissipated, is:$$W = fd = 13.72 \, \text{N} \times 5.6 \, \text{m} = 76.832 \, \text{J}$$Rounding to the nearest tenth gives 76.8 J, which corresponds to choice E.

Question 17

An astronaut tosses a ball out in space where gravitational forces may be neglected.What will happen to the ball?

Accepted Answer

In space, where gravitational forces can be neglected, an object in motion will continue in motion in a straight line at a constant speed according to Newton's first law of inertia, unless acted upon by an external force. The only significant external force that might act on the ball in the vacuum of space would be collisions with molecules or particles, which would very gradually slow it down.

Question 18

Objects A and B,of mass M and 2M respectively,are each pushed a distance d straight up an inclined plane by a force F parallel to the plane.The coefficient of kinetic friction between each mass and the plane has the same value $\mu$_k.At the highest point,

Accepted Answer

The answer of Objects A and B,of mass M and...

Question 19

Jane and Jake are looking at what happens to body 1 of mass m and body 2 of mass 2m,initially at rest,when equal forces are applied separately to the two bodies.Jake says that equal forces applied for equal times do equal amounts of work on the two bodies.Jane says that the two forces do equal amounts of work only if the two bodies move equal distances in the direction of the forces.Which one,if either,is correct?

Accepted Answer

Jane is correct because work is defined as force times distance, and the two forces are equal. Therefore, in order for the two forces to do equal amounts of work, the two bodies must move equal distances in the direction of the forces. The mass of a body does not affect the amount of work done on it, only the force and distance do. Therefore, Jake's statement is incorrect.

Question 20

At what rate is the gravitational force on a 2.0-kg projectile doing work at an instant when the velocity of the projectile is 4.0 m/s directed 30$\degree$ above the horizontal?

Accepted Answer

The gravitational force on the projectile is given by Fg = mg, where m is the mass of the projectile and g is the acceleration due to gravity. Therefore, Fg = (2.0 kg)(9.81 m/s^2) = 19.62 N.

At an instant when the velocity of the projectile is 4.0 m/s directed 30 degrees above the horizontal, the horizontal component of the velocity is v*cos(30) = (4.0 m/s)cos(30) = 3.46 m/s and the vertical component of the velocity is v*sin(30) = (4.0 m/s)sin(30) = 2.00 m/s.

The work done by the gravitational force is given by W = Fg*d*cos(theta), where d is the distance traveled by the projectile and theta is the angle between the force and the displacement. Since the projectile is moving in a parabolic path, d is changing continuously and is not well-defined at an instant. However, at this instant, the force is acting vertically downward (theta = 90 degrees) and the displacement is vertical as well (d = 0). Therefore, no work is being done by the gravitational force on the projectile at this instant (W = 0), and the rate of work is also zero (P = W/t = 0).

Thus, the only answer choice that is reasonable is C, which is negative and corresponds to the rate of work that would be done by a resisting force that is opposing the motion of the projectile. However, since no such force is mentioned in the problem, we cannot choose any of the answer choices with confidence.

Quiz 7: Conservation of Energy

Objects A and B,of mass M and 2M respectively,are each pushed a distance d straight up an inclined plane by a force F parallel to the plane.The coefficient of kinetic friction between each mass and the plane has the same value $\mu$ _k.At the highest point,

An all-terrain vehicle of 2 000 kg mass moves up a 15.0 $\degree$ slope at a constant velocity of 6.00 m/s.The rate of change of gravitational potential energy with time is

A champion athlete can produce one horsepower (746 W) for a short period of time.If a 70-kg athlete were to bicycle to the summit of a 500-m high mountain while expending power at this rate,she would have used at least ____ J of energy.

As a result of friction between internal parts of an isolated system

A 2.0-kg mass is projected vertically upward from ground level with an initial speed of 30 m/s.The mass rises to a maximum height of 35 m above ground level.What is the change in mechanical energy of the mass caused by air resistance between the point of projection and the point of maximum height?

A 6.0-kg block slides along a horizontal surface.If µ_k = 0.20 for the block and surface,at what rate is the friction force changing the mechanical energy of the block at an instant when its speed is 4.0 m/s?

As an object moves from point A to point B only two forces act on it: one force is conservative and does -70 J of work,the other force is nonconservative and does +50 J of work.Between A and B,

A 2.0-kg block is projected down a plane that makes an angle of 20 $\degree$ with the horizontal with an initial kinetic energy of 2.0 J.If the coefficient of kinetic friction between the block and plane is 0.40,how far will the block slide down the plane before coming to rest?

A champion athlete can produce one horsepower (746 W) for a short period of time.If a 70-kg athlete were to bicycle to the summit of a 500-m high mountain while expending power at this rate,she would reach the summit in ____ seconds.

As an object moves from point A to point B only two forces act on it: one force is nonconservative and does -30 J of work,the other force is conservative and does +50 J of work.Between A and B,

The two masses in the figure are released from rest.After the 3.0-kg mass has fallen 1.5 m,it is moving with a speed of 3.8 m/s.What is the change in mechanical energy done on the system during this time interval by the frictional force on the 2.0 kg mass?

An astronaut tosses a ball out in space where gravitational forces may be neglected.What will happen to the ball?

Objects A and B,of mass M and 2M respectively,are each pushed a distance d straight up an inclined plane by a force F parallel to the plane.The coefficient of kinetic friction between each mass and the plane has the same value $\mu$ _k.At the highest point,

At what rate is the gravitational force on a 2.0-kg projectile doing work at an instant when the velocity of the projectile is 4.0 m/s directed 30 $\degree$ above the horizontal?

Quiz 7: Conservation of Energy

Objects A and B,of mass M and 2M respectively,are each pushed a distance d straight up an inclined plane by a force F parallel to the plane.The coefficient of kinetic friction between each mass and the plane has the same value μ\muμ k.At the highest point,

An all-terrain vehicle of 2 000 kg mass moves up a 15.0 °\degree° slope at a constant velocity of 6.00 m/s.The rate of change of gravitational potential energy with time is

A champion athlete can produce one horsepower (746 W) for a short period of time.If a 70-kg athlete were to bicycle to the summit of a 500-m high mountain while expending power at this rate,she would have used at least ____ J of energy.

As a result of friction between internal parts of an isolated system

A 2.0-kg mass is projected vertically upward from ground level with an initial speed of 30 m/s.The mass rises to a maximum height of 35 m above ground level.What is the change in mechanical energy of the mass caused by air resistance between the point of projection and the point of maximum height?

A 6.0-kg block slides along a horizontal surface.If µk = 0.20 for the block and surface,at what rate is the friction force changing the mechanical energy of the block at an instant when its speed is 4.0 m/s?

As an object moves from point A to point B only two forces act on it: one force is conservative and does -70 J of work,the other force is nonconservative and does +50 J of work.Between A and B,

A 2.0-kg block is projected down a plane that makes an angle of 20 °\degree° with the horizontal with an initial kinetic energy of 2.0 J.If the coefficient of kinetic friction between the block and plane is 0.40,how far will the block slide down the plane before coming to rest?

A champion athlete can produce one horsepower (746 W) for a short period of time.If a 70-kg athlete were to bicycle to the summit of a 500-m high mountain while expending power at this rate,she would reach the summit in ____ seconds.

As an object moves from point A to point B only two forces act on it: one force is nonconservative and does -30 J of work,the other force is conservative and does +50 J of work.Between A and B,

The two masses in the figure are released from rest.After the 3.0-kg mass has fallen 1.5 m,it is moving with a speed of 3.8 m/s.What is the change in mechanical energy done on the system during this time interval by the frictional force on the 2.0 kg mass?

An astronaut tosses a ball out in space where gravitational forces may be neglected.What will happen to the ball?

Objects A and B,of mass M and 2M respectively,are each pushed a distance d straight up an inclined plane by a force F parallel to the plane.The coefficient of kinetic friction between each mass and the plane has the same value μ\muμ k.At the highest point,

At what rate is the gravitational force on a 2.0-kg projectile doing work at an instant when the velocity of the projectile is 4.0 m/s directed 30 °\degree° above the horizontal?

Objects A and B,of mass M and 2M respectively,are each pushed a distance d straight up an inclined plane by a force F parallel to the plane.The coefficient of kinetic friction between each mass and the plane has the same value $\mu$ _k.At the highest point,

An all-terrain vehicle of 2 000 kg mass moves up a 15.0 $\degree$ slope at a constant velocity of 6.00 m/s.The rate of change of gravitational potential energy with time is

A 6.0-kg block slides along a horizontal surface.If µ_k = 0.20 for the block and surface,at what rate is the friction force changing the mechanical energy of the block at an instant when its speed is 4.0 m/s?

A 2.0-kg block is projected down a plane that makes an angle of 20 $\degree$ with the horizontal with an initial kinetic energy of 2.0 J.If the coefficient of kinetic friction between the block and plane is 0.40,how far will the block slide down the plane before coming to rest?

Objects A and B,of mass M and 2M respectively,are each pushed a distance d straight up an inclined plane by a force F parallel to the plane.The coefficient of kinetic friction between each mass and the plane has the same value $\mu$ _k.At the highest point,

At what rate is the gravitational force on a 2.0-kg projectile doing work at an instant when the velocity of the projectile is 4.0 m/s directed 30 $\degree$ above the horizontal?