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AP Physics 1 : Gravitational Potential Energy

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Example Questions

Example Question #31 : Gravitational Potential Energy

Rollercoaster

The train car on the rollercoaster above has a mass of $726\ kg$ , and the potential energy at the top of the first hill is $292\ kJ$ .

What is the height of the hill?

Possible Answers:

$81\ m$

$31\ m$

$41\ m$

$25\ m$

$67\ m$

Correct answer:

$41\ m$

Explanation:

Potential energy is equal to the mass of an object multiplied by gravity multiplied by the height at which the object rests. This is demonstrated by the following equation:

PE=mgh

In this problem, you are given the mass and potential energy of the object. Gravity is a constant, therefore there is a single variable left to solve for and this is the height at which the object rests. By plugging in the given values the height is found to be:

292,000=726*9.81*h

$h\approx41\ m$

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Example Question #31 : Gravitational Potential Energy

A specialized hot air balloon is 1km high. Determine the initial velocity needed for a supply package to be launched straight up to reach the balloon. Ignore air resistance.

Possible Answers:

$9.81\frac{m}{s}$

$196\frac{m}{s}$

$140\frac{m}{s}$

$9800\frac{m}{s}$

None of these

Correct answer:

$140\frac{m}{s}$

Explanation:

1km=1000m

The package will need to at least gain 1000m in height, thus it will need to gain

$1000m*9.8\frac{m}{s^2}*mass$ gravitational potential energy

When it is launched, it will have a kinetic energy equal to

.5*mass*v^2

Since energy is conserved:

1000*9.8*m=.5mv^2

Solving for

$v=140\frac{m}{s}$

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Example Question #31 : Gravitational Potential Energy

A ball with a mass of 2kg is released from a ramp with a height of and the ball starts to roll downward. What is the velocity of the ball at the bottom of the ramp? (Assume no friction or air resistance).

Possible Answers:

$100\frac{m}{s}$

$150\frac{m}{s}$

$50\frac{m}{s}$

$75\frac{m}{s}$

$5\frac{m}{s}$

Correct answer:

$5\frac{m}{s}$

Explanation:

All of the potential energy that the ball has at the top of the ramp is converted into kinetic energy. Knowing this, we can determine the velocity when the ball is moving at the bottom of the ramp by assuming all the potential energy is converted into kinetic energy.

$mgh = \frac{1}{2}mv^{2}$

$mgh=2kg(10m}{s^2})(5m})$

$mgh=100J=\frac{1}{2}mv^2$

50J=2kg(v^2)

$v=5\frac{m}{s}$

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Example Question #31 : Gravitational Potential Energy

A jet is flying at $100\frac{m}{s}$ , 100m above the ocean. A bowling ball is dropped from the jet. At what velocity does the ball strike the water? Air resistance may be ignored.

Possible Answers:

$110\frac{m}{s}$

$100\frac{m}{s}$

None of these

$90\frac{m}{s}$

$144\frac{m}{s}$

Correct answer:

$110\frac{m}{s}$

Explanation:

The ball will gain velocity in the vertical direction due to gravity.

$v_y=\sqrt{2gh}$

The ball will also have velocity due to it's initial horizontal velocity

v_x=100

using

$v=\sqrt{v_x^2+v_y^2}$

Combining equations and plugging in values:

$v=\sqrt{100^2+2*9.8*100}$

$v=110\frac{m}{s}$

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AP Physics 1 : Gravitational Potential Energy

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #31 : Gravitational Potential Energy

Example Question #31 : Gravitational Potential Energy

Example Question #31 : Gravitational Potential Energy

Example Question #31 : Gravitational Potential Energy

All AP Physics 1 Resources

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