Question 1

A baseball outfielder throws a baseball of mass 0.36 kg at a speed of 34.0 m/s and initial angle of 30°.What is the kinetic energy of the baseball at the highest point of the trajectory? Ignore air friction.

Accepted Answer

At the highest point of the trajectory, the vertical component of the velocity is zero, but the horizontal component remains unchanged. The kinetic energy (KE) at the highest point is due to the horizontal component of the velocity only. The initial velocity can be broken down into horizontal (v_x) and vertical (v_y) components using trigonometry. Given the angle (θ) is 30° and the speed (v) is 34.0 m/s:$$v_x = v \cdot \cos(\theta) = 34.0 \cdot \cos(30°) = 34.0 \cdot \frac{\sqrt{3}}{2} \approx 29.4 \, \text{m/s}$$The kinetic energy is given by the formula:$$KE = \frac{1}{2}mv^2$$Substituting the mass (m = 0.36 kg) and the horizontal velocity (v_x ≈ 29.4 m/s) into the formula gives:$$KE = \frac{1}{2} \cdot 0.36 \cdot (29.4)^2 \approx 156 \, \text{J}$$Therefore, the kinetic energy of the baseball at the highest point of its trajectory is approximately 156 J.

Question 2

A 60-kg (including the passenger)sled is subject to a net force of 20 N pushing in the direction of the sled's motion as it is moving over a horizontal surface for a distance of 11 m after having started from rest.At this point the sled is released as it starts down a 10.0° incline.However,the snow is not very deep,and the sled stops after having moved an additional 35 m.What is the work done by friction while the sled is on the incline?

Accepted Answer

The answer of A 60-kg (including the passenger)sled is subject...

Question 3

A simple pendulum,1.00 m in length,is released from rest when the support string is at an angle of 48.0° from the vertical.What is the speed of the suspended mass at the bottom of the swing? (g = 9.80 m/s² and ignore air resistance)

Accepted Answer

The answer of A simple pendulum,1.00 m in length,is released...

Question 4

Old Faithful geyser in Yellowstone Park shoots water hourly to a height of 40.0 m.With what velocity does the water leave the ground?

Accepted Answer

The answer of Old Faithful geyser in Yellowstone Park shoots...

Question 5

A 3.00-kg ball has zero potential and kinetic energy.Maria drops the ball into a 10.0-m-deep well.After the ball comes to a stop in the mud,the sum of its potential and kinetic energy is:

Accepted Answer

When the ball is dropped into the well, it loses potential energy, which is converted into kinetic energy. Upon coming to a stop in the mud, all the kinetic energy is dissipated, but the ball has gained negative potential energy due to being below the reference point (where it had zero potential energy). The potential energy change is given by $PE = mgh$, where $m = 3.00 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, and $h = -10.0 \, \text{m}$ (negative because it's below the reference point). Thus, $PE = 3.00 \times 9.8 \times (-10.0) = -294 \, \text{J}$.

Question 6

An amount of work equal to 2.0 J is required to compress the spring in a spring-gun.What is the "launch speed" of a 26.0-g marble?

Accepted Answer

The launch speed can be found using the conservation of energy principle. The work done to compress the spring (2.0 J) is converted into the kinetic energy of the marble. The kinetic energy (KE) is given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass in kilograms and $v$ is the velocity in meters per second. Rearranging for $v$, we get $v = \sqrt{\frac{2KE}{m}}$. Substituting the given values ($KE = 2.0 J$, $m = 26.0 g = 0.026 kg$), we find $v = \sqrt{\frac{2 \times 2.0}{0.026}} \approx 12.4 m/s$.

Question 7

A pendulum consists of a mass m suspended from a string of length L.When the pendulum is swinging,how much greater is the mass's potential energy when the string is displaced from the vertical by the angle θ than when the mass is at the lowest position?

Accepted Answer

At the lowest position of the pendulum, the potential energy of the mass is given by:
PE = mgh
where h is the height of the mass from the lowest position, which is equal to L.

When the string is displaced from the vertical by an angle θ, the potential energy of the mass is given by:
PE = mgh'
where h' is the new height of the mass from the lowest position. Using trigonometry, we can find that:
h' = L(1-cosθ)

Substituting this value in the above equation for PE, we get:
PE = mgL(1-cosθ)

Therefore, the mass's potential energy is greater by mgL(1- cosθ) when the string is displaced from the vertical by the angle θ. Thus, the correct answer is D.

Question 8

A baseball catcher puts on an exhibition by catching a 0.175-kg ball dropped from a helicopter at a height of 89.0 m.What is the speed of the ball just before it hits the catcher's glove 1.00 m above the ground? (g = 9.80 m/s² and ignore air resistance)

Accepted Answer

The correct answer is A because the speed of the ball just before it hits the catcher's glove can be calculated using the conservation of energy principle. The potential energy of the ball at the height of 89.0 m is converted into kinetic energy just before it hits the catcher's glove. The kinetic energy of the ball is given by:$$KE = 1/2 mv^2$$where m is the mass of the ball and v is its speed. The potential energy of the ball at the height of 89.0 m is given by:$$PE = mgh$$where g is the acceleration due to gravity and h is the height. The conservation of energy principle states that the total energy of the system (PE + KE) is constant. Therefore, we have:$$PE = KE$$$$mgh = 1/2 mv^2$$Solving for v, we get:$$v = sqrt(2gh)$$Substituting the given values, we get:$$v = sqrt(2 * 9.80 m/s^2 * 89.0 m) = 41.5 m/s$$

Question 9

A 26.0-N crate starting at rest slides down a rough 4.8-m-long ramp,inclined at 21.0° with the horizontal.18 J of energy is lost to friction.What will be the speed of the crate at the bottom of the incline?

Accepted Answer

The answer of A 26.0-N crate starting at rest slides...

Question 10

An airliner with mass 90 000 kg is flying at 900 km/h at a height of 15.0 km.What is its total energy (kinetic + potential)if the total was 0 when the airliner was at rest on the ground?

Accepted Answer

The answer of An airliner with mass 90 000 kg...

Question 11

A girl and her bicycle have a total mass of 46.0 kg.At the top of the hill her speed is 6.0 m/s,and her speed doubles as she rides down the hill.The hill is 15.0 m high and 105 m long.How much kinetic energy and potential energy is lost to friction?

Accepted Answer

At the top of the hill, the total energy consists of both potential energy (due to height) and kinetic energy (due to speed). As she descends, some of this energy is converted into kinetic energy (due to increased speed), and some is lost to friction.1. Calculate initial potential energy (PE_initial) at the top:$$PE_{initial} = mgh = 46 \times 9.8 \times 15 = 6,774 \, J$$2. Calculate initial kinetic energy (KE_initial) at the top:$$KE_{initial} = \frac{1}{2}mv^2 = \frac{1}{2} \times 46 \times (6)^2 = 828 \, J$$3. Calculate total initial energy:$$E_{initial} = PE_{initial} + KE_{initial} = 6,774 + 828 = 7,602 \, J$$4. Calculate final kinetic energy (KE_final) when speed doubles (12 m/s):$$KE_{final} = \frac{1}{2}mv^2 = \frac{1}{2} \times 46 \times (12)^2 = 3,312 \, J$$5. Since she is at the bottom of the hill, potential energy (PE_final) is 0.6. Calculate total final energy:$$E_{final} = KE_{final} + PE_{final} = 3,312 + 0 = 3,312 \, J$$7. Calculate energy lost to friction:$$E_{lost} = E_{initial} - E_{final} = 7,602 - 3,312 = 4,290 \, J$$Therefore, the energy lost to friction is approximately 4.3E+3 J, which corresponds to choice A.

Question 12

An Olympic skier moving at 19.5 m/s down a 17.0° slope encounters a region of wet snow,of coefficient of friction μ_k = 0.740.How far down the slope does she go before stopping?

Accepted Answer

To find out how far down the slope she goes before stopping, we need to use the formula:

vf^2 = vi^2 + 2as

Where vf is the final velocity (0 m/s, since she stops), vi is the initial velocity (19.5 m/s), a is the acceleration (due to the force of friction), and s is the distance traveled.

We can find the acceleration using:

a = μ*g*cos(θ)

Where μ is the coefficient of friction, g is the acceleration due to gravity (9.81 m/s^2), and θ is the angle of the slope (17.0°):

a = 0.740*9.81*cos(17.0°) ≈ 1.459 m/s^2

Now we can use the first equation:

0 = (19.5 m/s)^2 + 2*(-1.459 m/s^2)*s

Solving for s, we get:

s ≈ 46.7 m

Therefore, the answer is B) 46.7 m.

Question 13

A box is launched up a frictionless incline with an angle of elevation θ.The maximum height reached by the box is h.The box is again launched up the incline with the same initial speed it had before,but this time the angle of elevation has been increased to φ.How does the maximum height H reached this time by the box compare with the previous value h?

Accepted Answer

The answer of A box is launched up a frictionless...

Question 14

A girl and her bicycle have a total mass of 42 kg.At the top of the hill her speed is 4.00 m/s.The hill is 14.3 m high and 112 m long.If the force of friction as she rides down the hill is 20 N,what is her speed at the bottom?

Accepted Answer

First, we need to determine the potential energy at the top of the hill:
PE = mgh = 42 kg * 9.81 m/s^2 * 14.3 m = 5802.18 J

Next, we need to find the work done by friction:
Wfriction = f * d = 20 N * 112 m = 2240 J

The total energy at the top of the hill equals the kinetic energy at the bottom of the hill plus the work done by friction:
PE = KE + Wfriction
KE = PE - Wfriction
KE = 5802.18 J - 2240 J = 3562.18 J

Finally, we can use the kinetic energy formula to find the velocity at the bottom of the hill:
KE = 1/2mv^2
v = sqrt((2KE)/m)
v = sqrt((2*3562.18 J)/(42 kg))
v ≈ 13.8 m/s

Therefore, the best choice is C, 13.8 m/s.

Question 15

A 2.50-kg projectile is launched at an angle of 20.0° above the horizontal with an initial speed of 100 m/s from the top of a building 40 m high.If the projectile is traveling 85.0 m/s at its maximum height of 47.0 m,how much work has been done on the projectile by air friction?

Accepted Answer

The work done by air friction can be found by calculating the difference in mechanical energy at the start and at the maximum height. Initially, the projectile has both kinetic and potential energy, and at its maximum height, it has less kinetic energy (due to air friction) but more potential energy. The initial kinetic energy (KE_initial) is given by $ \frac{1}{2} m v^2 $, where $ m = 2.50 \, \text{kg} $ and $ v = 100 \, \text{m/s} $, and the initial potential energy (PE_initial) is $ mgh $, where $ g = 9.8 \, \text{m/s}^2 $ and $ h = 40 \, \text{m} $. At the maximum height, the kinetic energy (KE_max) is $ \frac{1}{2} m v'^2 $, where $ v' = 85.0 \, \text{m/s} $, and the potential energy (PE_max) is $ mg(h + \Delta h) $, where $ \Delta h = 47.0 \, \text{m} - 40 \, \text{m} = 7.0 \, \text{m} $.Calculating the initial and maximum energies:- $ KE_{\text{initial}} = \frac{1}{2} \cdot 2.50 \cdot (100)^2 = 12500 \, \text{J} $- $ PE_{\text{initial}} = 2.50 \cdot 9.8 \cdot 40 = 980 \, \text{J} $- $ KE_{\text{max}} = \frac{1}{2} \cdot 2.50 \cdot (85.0)^2 = 5406.25 \, \text{J} $- $ PE_{\text{max}} = 2.50 \cdot 9.8 \cdot 47.0 = 1151.5 \, \text{J} $The total initial energy is $ 12500 + 980 = 13480 \, \text{J} $, and the total energy at the maximum height is $ 5406.25 + 1151.5 = 6557.75 \, \text{J} $. The work done by air friction is the difference, $ 13480 - 6557.75 = 6922.25 \, \text{J} $. However, since work done by air friction is energy lost, it should be negative, indicating an error in the calculation or interpretation. The correct approach involves calculating the change in mechanical energy correctly and accounting for the negative sign due to energy loss to friction. The provided choices suggest a calculation mistake or misunderstanding in the explanation process, as none of the calculations directly lead to the provided options. Re-evaluating the correct approach:The correct calculation involves finding the total mechanical energy at the beginning and at the maximum height, then finding the difference to determine the work done by air friction. The mistake in the explanation lies in the incorrect calculation or interpretation of the energy values and their differences. The correct answer should reflect the loss in mechanical energy due to air friction, which is not accurately captured in the provided explanation steps.

Question 16

A pole vaulter clears 3.98 m.With what speed does he strike the mat in the landing area?

Accepted Answer

The potential energy of the pole vaulter at the top of the jump is converted into kinetic energy just before the vaulter lands on the mat. Using conservation of energy,

mgh = (1/2)mv^2

where m is the mass of the pole vaulter, g is the acceleration due to gravity, h is the height the vaulter cleared, and v is the velocity just before landing on the mat. Solving for v gives:

v = sqrt(2gh)

Substituting the given values, we get:

v = sqrt(2*9.81*3.98) = 8.83 m/s

Therefore, the answer is C.

Question 17

A simple pendulum,2.6 m in length,is released with a push when the support string is at an angle of 45.0° from the vertical.If the initial speed of the suspended mass is 1.25 m/s when at the release point,what is its speed at the bottom of the swing? (g = 9.8 m/s²)

Accepted Answer

Using conservation of energy, we can relate the potential energy at the release point to the kinetic energy at the bottom of the swing:
mgh = (1/2)mv^2
where m is the mass of the suspended object, g is the acceleration due to gravity, h is the height above the bottom of the swing, and v is the speed at the bottom of the swing.
At the release point, the height above the bottom of the swing is h = L(1-cos(theta)) = 2.6(1-cos(45°)) = 1.68 m. Plugging in the given values, we get:
mgh = (1/2)mv^2
(4.4 kg)(9.8 m/s^2)(1.68 m) = (1/2)(4.4 kg)(1.25 m/s)^2 + (4.4 kg)v^2
Solving for v, we get:
v = sqrt[(2/7)(9.8)(1.68 - 0.78)] = 4.1 m/s.
Therefore, the answer is A.

Question 18

A bobsled makes a run down an ice track starting at 250 m vertical distance up the hill.If there is no friction,what is the velocity at the bottom of the hill?

Accepted Answer

The answer of A bobsled makes a run down an...

Question 19

When an object is dropped from a tower,what is the effect of the air resistance as it falls?

Accepted Answer

Air resistance acts to oppose the motion of the falling object, slowing it down and converting some of its kinetic energy into heat energy. However, this does not result in a positive work or an increase in potential energy. Therefore, none of the given choices are valid.

Question 20

A simple pendulum,2.0 m in length,is released by a push when the support string is at an angle of 25° from the vertical.If the initial speed of the suspended mass is 1.6 m/s when at the release point,to what maximum angle will it move in the second half of its swing?

Accepted Answer

The answer of A simple pendulum,2.0 m in length,is released...

Quiz 5: Energy

A baseball outfielder throws a baseball of mass 0.36 kg at a speed of 34.0 m/s and initial angle of 30°.What is the kinetic energy of the baseball at the highest point of the trajectory? Ignore air friction.

A simple pendulum,1.00 m in length,is released from rest when the support string is at an angle of 48.0° from the vertical.What is the speed of the suspended mass at the bottom of the swing? (g = 9.80 m/s² and ignore air resistance)

Old Faithful geyser in Yellowstone Park shoots water hourly to a height of 40.0 m.With what velocity does the water leave the ground?

A 3.00-kg ball has zero potential and kinetic energy.Maria drops the ball into a 10.0-m-deep well.After the ball comes to a stop in the mud,the sum of its potential and kinetic energy is:

An amount of work equal to 2.0 J is required to compress the spring in a spring-gun.What is the "launch speed" of a 26.0-g marble?

A pendulum consists of a mass m suspended from a string of length L.When the pendulum is swinging,how much greater is the mass's potential energy when the string is displaced from the vertical by the angle θ than when the mass is at the lowest position?

A baseball catcher puts on an exhibition by catching a 0.175-kg ball dropped from a helicopter at a height of 89.0 m.What is the speed of the ball just before it hits the catcher's glove 1.00 m above the ground? (g = 9.80 m/s² and ignore air resistance)

A 26.0-N crate starting at rest slides down a rough 4.8-m-long ramp,inclined at 21.0° with the horizontal.18 J of energy is lost to friction.What will be the speed of the crate at the bottom of the incline?

An airliner with mass 90 000 kg is flying at 900 km/h at a height of 15.0 km.What is its total energy (kinetic + potential) if the total was 0 when the airliner was at rest on the ground?

A girl and her bicycle have a total mass of 46.0 kg.At the top of the hill her speed is 6.0 m/s,and her speed doubles as she rides down the hill.The hill is 15.0 m high and 105 m long.How much kinetic energy and potential energy is lost to friction?

An Olympic skier moving at 19.5 m/s down a 17.0° slope encounters a region of wet snow,of coefficient of friction μ_k = 0.740.How far down the slope does she go before stopping?

A girl and her bicycle have a total mass of 42 kg.At the top of the hill her speed is 4.00 m/s.The hill is 14.3 m high and 112 m long.If the force of friction as she rides down the hill is 20 N,what is her speed at the bottom?

A 2.50-kg projectile is launched at an angle of 20.0° above the horizontal with an initial speed of 100 m/s from the top of a building 40 m high.If the projectile is traveling 85.0 m/s at its maximum height of 47.0 m,how much work has been done on the projectile by air friction?

A pole vaulter clears 3.98 m.With what speed does he strike the mat in the landing area?

A simple pendulum,2.6 m in length,is released with a push when the support string is at an angle of 45.0° from the vertical.If the initial speed of the suspended mass is 1.25 m/s when at the release point,what is its speed at the bottom of the swing? (g = 9.8 m/s²)

A bobsled makes a run down an ice track starting at 250 m vertical distance up the hill.If there is no friction,what is the velocity at the bottom of the hill?

When an object is dropped from a tower,what is the effect of the air resistance as it falls?

A simple pendulum,2.0 m in length,is released by a push when the support string is at an angle of 25° from the vertical.If the initial speed of the suspended mass is 1.6 m/s when at the release point,to what maximum angle will it move in the second half of its swing?

Quiz 5: Energy

A baseball outfielder throws a baseball of mass 0.36 kg at a speed of 34.0 m/s and initial angle of 30°.What is the kinetic energy of the baseball at the highest point of the trajectory? Ignore air friction.

A simple pendulum,1.00 m in length,is released from rest when the support string is at an angle of 48.0° from the vertical.What is the speed of the suspended mass at the bottom of the swing? (g = 9.80 m/s2 and ignore air resistance)

Old Faithful geyser in Yellowstone Park shoots water hourly to a height of 40.0 m.With what velocity does the water leave the ground?

A 3.00-kg ball has zero potential and kinetic energy.Maria drops the ball into a 10.0-m-deep well.After the ball comes to a stop in the mud,the sum of its potential and kinetic energy is:

An amount of work equal to 2.0 J is required to compress the spring in a spring-gun.What is the "launch speed" of a 26.0-g marble?

A pendulum consists of a mass m suspended from a string of length L.When the pendulum is swinging,how much greater is the mass's potential energy when the string is displaced from the vertical by the angle θ than when the mass is at the lowest position?

A baseball catcher puts on an exhibition by catching a 0.175-kg ball dropped from a helicopter at a height of 89.0 m.What is the speed of the ball just before it hits the catcher's glove 1.00 m above the ground? (g = 9.80 m/s2 and ignore air resistance)

A 26.0-N crate starting at rest slides down a rough 4.8-m-long ramp,inclined at 21.0° with the horizontal.18 J of energy is lost to friction.What will be the speed of the crate at the bottom of the incline?

An airliner with mass 90 000 kg is flying at 900 km/h at a height of 15.0 km.What is its total energy (kinetic + potential) if the total was 0 when the airliner was at rest on the ground?

A girl and her bicycle have a total mass of 46.0 kg.At the top of the hill her speed is 6.0 m/s,and her speed doubles as she rides down the hill.The hill is 15.0 m high and 105 m long.How much kinetic energy and potential energy is lost to friction?

An Olympic skier moving at 19.5 m/s down a 17.0° slope encounters a region of wet snow,of coefficient of friction μk = 0.740.How far down the slope does she go before stopping?

A girl and her bicycle have a total mass of 42 kg.At the top of the hill her speed is 4.00 m/s.The hill is 14.3 m high and 112 m long.If the force of friction as she rides down the hill is 20 N,what is her speed at the bottom?

A 2.50-kg projectile is launched at an angle of 20.0° above the horizontal with an initial speed of 100 m/s from the top of a building 40 m high.If the projectile is traveling 85.0 m/s at its maximum height of 47.0 m,how much work has been done on the projectile by air friction?

A pole vaulter clears 3.98 m.With what speed does he strike the mat in the landing area?

A simple pendulum,2.6 m in length,is released with a push when the support string is at an angle of 45.0° from the vertical.If the initial speed of the suspended mass is 1.25 m/s when at the release point,what is its speed at the bottom of the swing? (g = 9.8 m/s2)

A bobsled makes a run down an ice track starting at 250 m vertical distance up the hill.If there is no friction,what is the velocity at the bottom of the hill?

When an object is dropped from a tower,what is the effect of the air resistance as it falls?

A simple pendulum,2.0 m in length,is released by a push when the support string is at an angle of 25° from the vertical.If the initial speed of the suspended mass is 1.6 m/s when at the release point,to what maximum angle will it move in the second half of its swing?

A simple pendulum,1.00 m in length,is released from rest when the support string is at an angle of 48.0° from the vertical.What is the speed of the suspended mass at the bottom of the swing? (g = 9.80 m/s² and ignore air resistance)

A baseball catcher puts on an exhibition by catching a 0.175-kg ball dropped from a helicopter at a height of 89.0 m.What is the speed of the ball just before it hits the catcher's glove 1.00 m above the ground? (g = 9.80 m/s² and ignore air resistance)

An Olympic skier moving at 19.5 m/s down a 17.0° slope encounters a region of wet snow,of coefficient of friction μ_k = 0.740.How far down the slope does she go before stopping?

A simple pendulum,2.6 m in length,is released with a push when the support string is at an angle of 45.0° from the vertical.If the initial speed of the suspended mass is 1.25 m/s when at the release point,what is its speed at the bottom of the swing? (g = 9.8 m/s²)