Question 1

An arrow of mass 0.100 kg is launched directly upward with a bow of spring constant 1962 N/m.If the bow was stretched 10.0 cm and then released,how high above the bow did the arrow go before coming to rest temporarily?

Accepted Answer

The potential energy stored in the stretched bow is converted into gravitational potential energy at the highest point. The energy stored in the bow when stretched is given by $ \frac{1}{2}kx^2 $, where $ k $ is the spring constant and $ x $ is the stretch distance. Substituting $ k = 1962 \, \text{N/m} $ and $ x = 0.1 \, \text{m} $, we get $ \frac{1}{2} \times 1962 \times (0.1)^2 = 98.1 \, \text{J} $. This energy equals the gravitational potential energy at the highest point, given by $ mgh $, where $ m = 0.1 \, \text{kg} $, $ g = 9.8 \, \text{m/s}^2 $, and $ h $ is the height. Solving $ 98.1 = 0.1 \times 9.8 \times h $ gives $ h = 10.0 \, \text{m} $.

Question 2

A block of mass 1.40 kg is attached to a spring and sits on a frictionless table which is a height h=4.0 m above the floor.The spring is compressed by d=0.11 m initially.If the spring constant is k = 600 N/m,what is the speed of the block when it leaves the spring?

Accepted Answer

B

Question 3

Calculate the gravitational potential energy relative to the ground of a 35.0-kg rock is 1.75 m above the ground.

Accepted Answer

The formula for gravitational potential energy is given by:

PE = mgh

Where m = mass of the object, g = acceleration due to gravity, and h = height of the object from the ground.

Substituting the given values, we get:

PE = (35.0 kg)(9.81 m/s^2)(1.75 m) = 601 J

Therefore, the gravitational potential energy of the rock relative to the ground is 601 J. The correct answer is choice C.

Question 4

A 1210-kg car travels 1.20 km up an incline at constant velocity.The incline is 15S1U1P1$\circ$S1S1P0 measured with respect to the horizontal.The change in the car's potential energy is

Accepted Answer

The change in potential energy is given by mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the height of the incline. In this case, m = 1210 kg, g = 9.8 m/s^2, and h = 1.20 km * sin(15°) = 306 m. Therefore, the change in potential energy is:PE = mgh = (1210 kg)(9.8 m/s^2)(306 m) = 3.68 MJ

Question 5

The potential energy of an object as a function of position is U(x)= x³ - 12x + 8,where U is measured in joules and x is measured in meters.Where is the object be located if it is in an unstable equilibrium?

Accepted Answer

An unstable equilibrium occurs when a small displacement from an equilibrium position leads to an increase in potential energy. This can happen when the second derivative of the potential energy function is positive. In this case, the second derivative is U''(x) = 2³ + 24, which is positive when x is negative. Therefore, the object is located at x = -2.0 m.

Question 6

A 1210-kg car travels up an incline at constant velocity.The incline is 15S1U1P1$\circ$S1S1P0 measured with respect to the horizontal.The change in the car's potential energy is 1.92 MJ.What is the length of the incline?

Accepted Answer

The change in potential energy (ΔPE) can be calculated using the formula ΔPE = mgh, where m is the mass of the car, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height gained. Given ΔPE = 1.92 MJ (or 1.92 x 10^6 J) and m = 1210 kg, we can solve for h:1.92 x 10^6 J = 1210 kg * 9.8 m/s^2 * h,h = (1.92 x 10^6 J) / (1210 kg * 9.8 m/s^2) ≈ 160 m.To find the length of the incline (L), we use the trigonometric relation sin(θ) = opposite/hypotenuse, where θ = 10 degrees and the opposite side is the height (h). Thus,sin(10°) = h/L,L = h / sin(10°) ≈ 160 m / sin(10°) ≈ 920 m or 0.92 km.Among the given options, 0.62 km (E) is the closest to the calculated length of the incline.

Question 7

The potential energy of an object as a function of position is U(x)= x² - x - 6,where U is measured in joules and x is measured in meters.Where is the object be located if it is in a stable equilibrium?

Accepted Answer

To find the point of stable equilibrium, we need to look for the minimum of the potential energy function. 
Taking the derivative of the given function with respect to x, we get: 
U'(x) = S1U1P12S1S1P0 - 1 
Setting U'(x) = 0 to find the critical points, we get: 
S1U1P12S1S1P0 - 1 = 0 
Simplifying, we get: 
x = (S1U1P12S1S1P0 - 1)/S1S1P0 
Since U''(x) = -1 (the second derivative is negative), we can confirm that this critical point is a minimum, which is the location of the stable equilibrium. 
Plugging in the value of U1, we get: 
x = (S1S1P0 - 1)/S1S1P0 
x = (1/2)/1 
x = 1/2 
Therefore, the object is located at x = 0.5 m in a stable equilibrium.

Question 8

The gravitational potential energy of a 35.0-kg rock is 690.0 J relative to the ground.The rock is at a height ______ above the ground.

Accepted Answer

The formula for gravitational potential energy is U = mgh, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), h is the height above the reference level. 
Rearranging the formula to solve for height, we get h = U/mg.
Substituting the given values, h = (690.0 J) / (35.0 kg x 9.8 m/s^2) = 2.01 m (rounded to two decimal places). 
Therefore, the rock is at a height of 2.01 m above the ground. 
The best choice is C.

Question 9

A particle of mass m moving in the x-y plane is confined by a two-dimensional potential.The net force on the mass is  .What is the potential?

Accepted Answer

The answer of A particle of mass m moving in...

Question 10

An 80-kg fireman slides down a 3-meter pole by applying a frictional force against the pole using his hands.If he slides from rest and he land on the ground with a speed of 6.00 m/s,the frictional force his hands applied against the pole was

Accepted Answer

The answer of An 80-kg fireman slides down a 3-meter...

Question 11

A 30.0-kg crate is pushed across a rough floor by a horizontal force of 100 N.The coefficient of kinetic friction between the crate and the floor is 0.100.If a total of 80.5 J of work is done on the crate,how far was it moved along the floor?

Accepted Answer

The answer of A 30.0-kg crate is pushed across a...

Question 12

You are on a sled at the top of a 30.0-m tall toboggan run.Provided you can ignore friction forces between the sled and the track,when you reach the bottom of the toboggan you have a speed of

Accepted Answer

The speed of an object at the bottom of a hill is determined solely by the potential energy it had at the top of the hill. Using the equation for gravitational potential energy, PE = mgh, where m is the mass of the sled, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the hill (30.0 m), we can find the potential energy at the top of the hill.

PE = mgh 
PE = (m)(9.8 m/s^2)(30.0 m) 
PE = 2940 m kg/s^2

This potential energy is converted entirely into kinetic energy at the bottom of the hill. Using the equation for kinetic energy, KE = 1/2mv^2, where v is the speed of the sled at the bottom of the hill, we can solve for v.

KE = 1/2mv^2 
2940 m kg/s^2 = 1/2m(v^2) 
5880 m/s^2 = v^2 
v ≈ 24.3 m/s

Therefore, the answer is choice C.

Question 13

A stone is thrown up in the air and it reaches a maximum height of 10.0 m.What is the initial speed of the stone?

Accepted Answer

The initial speed can be calculated using the formula for maximum height in projectile motion: $ v^2 = u^2 - 2gh $. Here, $ v = 0 $ at maximum height, $ g = 9.8 \, \text{m/s}^2 $, and $ h = 10.0 \, \text{m} $. Solving for $ u $, we get $ u = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 10} = 14.0 \, \text{m/s} $.

Question 14

An Olympic weight lifter lifts a 186 kg weight from the floor to a height of 2.10 m.The change in the potential energy of the weight is

Accepted Answer

The potential energy (PE) formula is PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. 
Converting 186 kg to Newtons, we have F = ma = 186 kg x 9.81 m/s² = 1824.66 N. 
Work = force x distance, so W = (1824.66 N)(2.10 m) = 3832 J. 
The change in potential energy is equal to the work done on the weight, so the answer is D, 3832 J. 
Note that the negative signs in choices A and C are incorrect, as the weight is being lifted, so the potential energy is increasing, not decreasing. Choice E is not equal to the work done on the weight. Choice B is not the correct answer, as it only gives the work done against gravity, not the total work done.

Question 15

The potential energy of an object as a function of position is U(x)= x³ + 8,where U is measured in joules and x is measured in meters.Where is the object be located if it is in a metastable equilibrium?

Accepted Answer

For an object to be in metastable equilibrium, the first derivative of its potential energy with respect to position (dU/dx) should be zero, and the second derivative (d²U/dx²) should be greater than zero. The given potential energy function is U(x) = x^13 + 8. The first derivative with respect to x is dU/dx = 13x^12. Setting this equal to zero, we find that x = 0 is the only point where the derivative is zero, indicating a potential equilibrium point. The second derivative is d²U/dx² = 156x^11, which is positive for all x except x = 0, where it is exactly zero. However, since there's no other point where the first derivative is zero, x = 0 is the closest condition to a metastable equilibrium given the options, even though strictly speaking, a positive second derivative would be preferred to confirm stability.

Question 16

A loaded wheelbarrow with a total mass of 70 kg is moved from the bottom to the top of a 2.5-m tall hill.The change in the potential energy of the loaded wheelbarrow is

Accepted Answer

The answer of A loaded wheelbarrow with a total mass...

Question 17

A slingshot is used to launch a 5.00-gram rock.The effective spring constant of the slingshot is 200 N/m.By what distance must the slingshot be stretched in order to launch the rock at 20.0 m/s?

Accepted Answer

The potential energy stored in the slingshot is given by U = (1/2)kx^2, where k is the spring constant and x is the distance stretched. We can rearrange this equation to solve for x:
x = sqrt(2U/k)
To launch the rock at 20.0 m/s, the initial kinetic energy of the rock must be (1/2)mv^2 = 0.5 * 0.005 kg * (20.0 m/s)^2 = 2 J. This is also the amount of potential energy that must be stored in the slingshot before launch. Therefore:
x = sqrt(2 * 2 J / 200 N/m) = 0.1 m = 10 cm
Therefore, the slingshot must be stretched by 10 cm (or 0.1 m) to launch the rock at 20.0 m/s. The correct answer is C.

Question 18

A projectile of mass 2.00 kg is launched from the ground with a speed 10.0 m/s at an angle 20.0S1U1P1$\circ$S1S1P0 above the horizontal.Assume that air drag is negligible.What is the kinetic energy of the projectile 0.500 seconds after it is launched?

Accepted Answer

The answer of A projectile of mass 2.00 kg is...

Question 19

441 J of work is required to slide a 15-kg ice block up a frictionless slope that is 20 m long.The incline of the slope with respect to horizontal is

Accepted Answer

The work done to slide the ice block up the slope is equal to the gravitational potential energy gained, which is $mgh$, where $m$ is the mass, $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$), and $h$ is the height gained. Since the slope is frictionless, all the work goes into increasing the gravitational potential energy.Given that the work done $W = 441 \, \text{J}$, and the mass $m = 15 \, \text{kg}$, we can express the height gained as $h = \frac{W}{mg}$.First, we find $h$:$$h = \frac{441}{15 \times 9.8} = \frac{441}{147} = 3 \, \text{m}$$Knowing the length of the slope ($l = 20 \, \text{m}$) and the height ($h = 3 \, \text{m}$), we can find the sine of the angle ($\theta$) of the slope using the right triangle formed by the slope, height, and the horizontal ground:$$\sin(\theta) = \frac{h}{l} = \frac{3}{20}$$To find the angle $\theta$, we take the inverse sine ($\sin^{-1}$) of $\frac{3}{20}$:$$\theta = \sin^{-1}\left(\frac{3}{20}\right) \approx 8.62^\circ$$Therefore, the incline of the slope with respect to the horizontal is approximately $8.62^\circ$.

Question 20

The mechanical energy of an object is always equal to

Accepted Answer

Mechanical energy is the sum of an object's kinetic and potential energies, as given by the equation E = K + U. Work done on an object and change in kinetic energy are ways to change the mechanical energy, but not necessarily equal to it.

Quiz 6: Potential Energy and Energy Conservation

An arrow of mass 0.100 kg is launched directly upward with a bow of spring constant 1962 N/m.If the bow was stretched 10.0 cm and then released,how high above the bow did the arrow go before coming to rest temporarily?

A block of mass 1.40 kg is attached to a spring and sits on a frictionless table which is a height h=4.0 m above the floor.The spring is compressed by d=0.11 m initially.If the spring constant is k = 600 N/m,what is the speed of the block when it leaves the spring?

Calculate the gravitational potential energy relative to the ground of a 35.0-kg rock is 1.75 m above the ground.

A 1210-kg car travels 1.20 km up an incline at constant velocity.The incline is 15^$\circ$ measured with respect to the horizontal.The change in the car's potential energy is

The potential energy of an object as a function of position is U(x) = x³ - 12x + 8,where U is measured in joules and x is measured in meters.Where is the object be located if it is in an unstable equilibrium?

A 1210-kg car travels up an incline at constant velocity.The incline is 15^$\circ$ measured with respect to the horizontal.The change in the car's potential energy is 1.92 MJ.What is the length of the incline?

The potential energy of an object as a function of position is U(x) = x² - x - 6,where U is measured in joules and x is measured in meters.Where is the object be located if it is in a stable equilibrium?

The gravitational potential energy of a 35.0-kg rock is 690.0 J relative to the ground.The rock is at a height ______ above the ground.

A particle of mass m moving in the x-y plane is confined by a two-dimensional potential.The net force on the mass is .What is the potential?

An 80-kg fireman slides down a 3-meter pole by applying a frictional force against the pole using his hands.If he slides from rest and he land on the ground with a speed of 6.00 m/s,the frictional force his hands applied against the pole was

A 30.0-kg crate is pushed across a rough floor by a horizontal force of 100 N.The coefficient of kinetic friction between the crate and the floor is 0.100.If a total of 80.5 J of work is done on the crate,how far was it moved along the floor?

You are on a sled at the top of a 30.0-m tall toboggan run.Provided you can ignore friction forces between the sled and the track,when you reach the bottom of the toboggan you have a speed of

A stone is thrown up in the air and it reaches a maximum height of 10.0 m.What is the initial speed of the stone?

An Olympic weight lifter lifts a 186 kg weight from the floor to a height of 2.10 m.The change in the potential energy of the weight is

The potential energy of an object as a function of position is U(x) = x³ + 8,where U is measured in joules and x is measured in meters.Where is the object be located if it is in a metastable equilibrium?

A loaded wheelbarrow with a total mass of 70 kg is moved from the bottom to the top of a 2.5-m tall hill.The change in the potential energy of the loaded wheelbarrow is

A slingshot is used to launch a 5.00-gram rock.The effective spring constant of the slingshot is 200 N/m.By what distance must the slingshot be stretched in order to launch the rock at 20.0 m/s?

A projectile of mass 2.00 kg is launched from the ground with a speed 10.0 m/s at an angle 20.0^$\circ$ above the horizontal.Assume that air drag is negligible.What is the kinetic energy of the projectile 0.500 seconds after it is launched?

441 J of work is required to slide a 15-kg ice block up a frictionless slope that is 20 m long.The incline of the slope with respect to horizontal is

The mechanical energy of an object is always equal to

Quiz 6: Potential Energy and Energy Conservation

An arrow of mass 0.100 kg is launched directly upward with a bow of spring constant 1962 N/m.If the bow was stretched 10.0 cm and then released,how high above the bow did the arrow go before coming to rest temporarily?

A block of mass 1.40 kg is attached to a spring and sits on a frictionless table which is a height h=4.0 m above the floor.The spring is compressed by d=0.11 m initially.If the spring constant is k = 600 N/m,what is the speed of the block when it leaves the spring?

Calculate the gravitational potential energy relative to the ground of a 35.0-kg rock is 1.75 m above the ground.

A 1210-kg car travels 1.20 km up an incline at constant velocity.The incline is 15 ∘\circ∘ measured with respect to the horizontal.The change in the car's potential energy is

The potential energy of an object as a function of position is U(x) = x3 - 12x + 8,where U is measured in joules and x is measured in meters.Where is the object be located if it is in an unstable equilibrium?

A 1210-kg car travels up an incline at constant velocity.The incline is 15 ∘\circ∘ measured with respect to the horizontal.The change in the car's potential energy is 1.92 MJ.What is the length of the incline?

The potential energy of an object as a function of position is U(x) = x2 - x - 6,where U is measured in joules and x is measured in meters.Where is the object be located if it is in a stable equilibrium?

The gravitational potential energy of a 35.0-kg rock is 690.0 J relative to the ground.The rock is at a height ______ above the ground.

A particle of mass m moving in the x-y plane is confined by a two-dimensional potential.The net force on the mass is .What is the potential?

An 80-kg fireman slides down a 3-meter pole by applying a frictional force against the pole using his hands.If he slides from rest and he land on the ground with a speed of 6.00 m/s,the frictional force his hands applied against the pole was

A 30.0-kg crate is pushed across a rough floor by a horizontal force of 100 N.The coefficient of kinetic friction between the crate and the floor is 0.100.If a total of 80.5 J of work is done on the crate,how far was it moved along the floor?

You are on a sled at the top of a 30.0-m tall toboggan run.Provided you can ignore friction forces between the sled and the track,when you reach the bottom of the toboggan you have a speed of

A stone is thrown up in the air and it reaches a maximum height of 10.0 m.What is the initial speed of the stone?

An Olympic weight lifter lifts a 186 kg weight from the floor to a height of 2.10 m.The change in the potential energy of the weight is

The potential energy of an object as a function of position is U(x) = x3 + 8,where U is measured in joules and x is measured in meters.Where is the object be located if it is in a metastable equilibrium?

A loaded wheelbarrow with a total mass of 70 kg is moved from the bottom to the top of a 2.5-m tall hill.The change in the potential energy of the loaded wheelbarrow is

A slingshot is used to launch a 5.00-gram rock.The effective spring constant of the slingshot is 200 N/m.By what distance must the slingshot be stretched in order to launch the rock at 20.0 m/s?

A projectile of mass 2.00 kg is launched from the ground with a speed 10.0 m/s at an angle 20.0 ∘\circ∘ above the horizontal.Assume that air drag is negligible.What is the kinetic energy of the projectile 0.500 seconds after it is launched?

441 J of work is required to slide a 15-kg ice block up a frictionless slope that is 20 m long.The incline of the slope with respect to horizontal is

The mechanical energy of an object is always equal to

A 1210-kg car travels 1.20 km up an incline at constant velocity.The incline is 15^$\circ$ measured with respect to the horizontal.The change in the car's potential energy is

The potential energy of an object as a function of position is U(x) = x³ - 12x + 8,where U is measured in joules and x is measured in meters.Where is the object be located if it is in an unstable equilibrium?

A 1210-kg car travels up an incline at constant velocity.The incline is 15^$\circ$ measured with respect to the horizontal.The change in the car's potential energy is 1.92 MJ.What is the length of the incline?

The potential energy of an object as a function of position is U(x) = x² - x - 6,where U is measured in joules and x is measured in meters.Where is the object be located if it is in a stable equilibrium?

The potential energy of an object as a function of position is U(x) = x³ + 8,where U is measured in joules and x is measured in meters.Where is the object be located if it is in a metastable equilibrium?

A projectile of mass 2.00 kg is launched from the ground with a speed 10.0 m/s at an angle 20.0^$\circ$ above the horizontal.Assume that air drag is negligible.What is the kinetic energy of the projectile 0.500 seconds after it is launched?