AP Physics 1 - Gravitational Potential Energy

On the surface of planet Mars, the gravitational constant is $g_M=3.75 \frac{m}{s^2}$ . Considering that earth's gravitational constant is $g_E= 10 \frac{m}{s^2}$ , how high above the surface do you need to be on planet Mars to have the same gravitational potential energy as you would if you were up on Earth? Assume your mass is unchanged on both planets.

Explanation

Recall that the formula for gravitational potential energy is:

To determine the height on planet Mars needed to have equivalent potential energy, we can set the two equations for gravitational potential energies equal.

, where is mass and is height above the ground on Earth while is the height above the ground on Mars.

The mass, can be eliminated from both sides of the equation since they are equal. Plug in and solve for the height on Mars.

$10\frac{m}{s^2}*10m=3.75\frac{m}{s^2}*h_M$

$h_M= \frac{100}{3.75}m= 26.7m$

A roller coaster of mass is approaching a circular loop with a radius of . The average frictional force on the coaster is and the velocity of the coaster is $25\frac{m}{s}$ as it enters the loop. How many g's are felt by the riders of the coaster when it reaches the top of the loop?

Explanation

There are two forces acting on the coaster at the top of the loop: gravity and centrifugal. It is important to note in this problem the difference between centripetal and centrifugal forces. Centripetal force the is actual force applied to the coaster (points inward) and the centrifugal force is that apparent force felt by the coaster as it travels in a loop (points outward). Just remember that these forces are equal, but in opposite directions. Therefore, we will be using the centrifugal force to calculate the g's felt by the coaster. To do this, we will begin with the expression for conservation of energy:

If we assume that the bottom of the loop has a height of zero, we can eliminate initial potential energy. Also, the only work is done by friction, and it is work done on the surrounds, so it will be negative. Therefore, we get:

Plugging in expressions for each of these, we get:

$\frac{1}{2}mv_i^2+F_Fd=mgh_f+\frac{1}{2}mv_f^2$

Rearranging for final velocity, we get:

$v_f = \sqrt{\frac{mv_i^2-2F_Fd-2mgh_f}{m}}$

Where d is half the circumference of the loop:

$d=\pi r$

And the final height is twice the radius:

Plugging these in, we get:

$v_f = \sqrt{\frac{mv_i^2-2F_F\pi r-4mgr}{m}}$

Plugging in our values, we get:

$v_f = \sqrt{\frac{(1500kg)\left ( 25\frac{m}{s}\right )^2-2(600N)\pi (10m)-4(1500kg)\left ( 10\frac{m}{s^2}\right )(10m)}{1500kg}}$

$v_f = 14.14\frac{m}{s}$

Now we can use this to calculate the centrifugal force:

$F_c = ma_c=m\frac{v^2}{r}$

This is in the opposite direction of the other force acting on the system, gravity. Therefore, we get:

$F_{net}=F_c-F_g$

Plugging in expressions we get:

$F_{net}=m\frac{v^2}{r}-mg=m\left ( \frac{v^2}{r}-g\right )$

Rearranging for net acceleration, we get:

$F_{net}=ma=m\left ( \frac{v^2}{r}-g\right )$

$a=\left ( \frac{v^2}{r}-g\right )= 9.99\frac{m}{s^2} = 10\frac{m}{s^2}$

Then calculating g's we get:

$g's=\frac{a}{g}=\frac{10}{10}= 1g$

A ball rolls down a frictionless ramp of height , at the end of the ramp what will its velocity be?

$10\frac{m}{s}$

$5\frac{m}{s}$

$20\frac{m}{s}$

$2\frac{m}{s}$

Explanation

The first step for this problem is to determine the potential energy the ball has at the top of the ramp through this equation:

$U=m*g*h= (2kg)(10\frac{m}{s^2})(5m)= 100J$

We see that it has a potential energy of 100 joules. All of this potential energy gets converted to kinetic energy as the ball falls down the ramp.

Knowing this, we can determine the velocity the ball has at the bottom of the ramp by setting the potential energy equal to the kinetic energy:

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

We substitute the known mass and solve for :

$100J=\frac{1}{2}(2kg)v^2$

$v^2= 100\frac{m^2}{s^2}$

$v=10\frac{m}{s}$

$G=6.67*10^{-11}\frac{N*m^2}{kg^2}$

Mass of Pluto: $1.31*10^{22}kg$

Radius of Pluto: $1.19*10^{6}m$

Determine the escape velocity for a lander on Pluto.

$1.21*10^{3}\frac{m}{s}$

None of these

$4.22*10^{3}\frac{m}{s}$

$3.95*10^{3}\frac{m}{s}$

$1.05*10^{3}\frac{m}{s}$

Explanation

Using:

$-G\frac{m_1m_2}{r}=PE_{grav}$

The escape velocity is the velocity necessary to have a kinetic energy that can overcome the gravitational potential. That is, the kinetic energy plus the "big G" gravitational potential energy need to add to zero or a positive number.

$.5m_{lander}v_{lander}^2-G\frac{m_{lander}m_{pluto}}{r_{pluto}}=0$

Solving for $v_{lander}$

$v_{lander}=\sqrt(\frac{2*G*m_{pluto}}{r_{pluto}})$

Plugging in values:

$v_{lander}=\sqrt(\frac{2*6.67*10^{-11}*1.31*10^{22}}{1.19*10^{6}})$

$v_{lander}=1.21*10^{3}\frac{m}{s}$

Changing which of the following variables will cause the biggest increase in gravitational potential energy?

More than one of these will cause biggest increase

Distance between the two objects

Mass of the object

Velocity

Explanation

Gravitational potential energy is calculated using the following equation.

$U = \frac{-GMm}{r}$

Where is the potential energy, is the gravitational constant, is the mass of one object, is the mass of the other object, and is the distance between the two objects. Decreasing and increasing mass of the object by the same factor will have a similar effect on the potential energy (both will increase potential energy by the same factor); therefore, changing both of these variables by same factor will have a similar effect. Recall that the force due to gravity is

$F_g = \frac{-GMm}{r^2}$

When calculating , will have the biggest effect. Don't confuse equation for potential energy with equation for force due to gravity.

Consider the following system:

Spinning rod with masses at end

Two spherical masses, A and B, are attached to the end of a rigid rod with length l. The rod is attached to a fixed point, p, which is at the midpoint between the masses and is at a height, h, above the ground. The rod spins around the fixed point in a vertical circle that is traced in grey. $\angle c$ is the angle at which the rod makes with the horizontal at any given time ( $c=0^{\circ}$ in the figure).

The rod is in it's vertical position (mass A is at the top, mass B is at the bottom). If mass A has twice as much gravitational potential energy as mass B, what is h?

$g = 10\frac{m}{s^2}$

Explanation

We will only use the expression for gravitational potential energy to solve this problem:

Written out for each mass:

We are asked to find at which point does mass A have twice as much gravitational potential energy as mass B:

$\frac{U_A}{U_B}=2$

$2=\frac{m_Agh_A}{m_Bgh_B}=\frac{m_Ah_A}{m_Bh_B}$

Now we need to determine the heights of each mass which are the height given in the problem +/- half the length of the rod:

$h_A=h-\frac{1}{2}l$

$h_B=h+\frac{1}{2}l$

Plugging these into our expression, we get:

$2=\frac{m_A\left ( h+\frac{1}{2}l\right )}{m_B\left ( h-\frac{1}{2}l\right )}$

We know all the values in this expression, so we just need to isolate h:

$2m_B\left ( h-\frac{1}{2}l\right )=m_A\left ( h+\frac{1}{2}l\right )$

Expanding both sides:

$2m_Bh - m_Bl = m_Ah + \frac{1}{2}m_Al$

Rearranging to get h on the same side:

$2m_Bh-m_Ah=m_Bl+\frac{1}{2}m_Al$

Factoring out h and rearranging:

$h(2m_B-m_A) = m_Bl+\frac{1}{2}m_Al$

$h=\frac{m_Bl+\frac{1}{2}l}{2m_B-m_A}$

This is our final equation, time to plug and chug:

$h = \frac{(20kg)(6m)+\frac{1}{2}(30kg)(6m)}{2(20kg)-30kg} = \frac{120kg\cdot m+90kg\cdot m}{40kg-30kg}$

$h = \frac{210kg\cdot m}{10kg}=21m$

A mountain biker goes off a jump with an initial vertical velocity of $8\frac{m}{s}$ . If the biker lands a vertical distance of below the launch point, what is his vertical velocity the moment he lands?

$g = 10\frac{m}{s^2}$

$11.1\frac{m}{s}$

$15.4\frac{m}{s}$

$8.6\frac{m}{s}$

$10.2\frac{m}{s}$

$9.9\frac{m}{s}$

Explanation

We can use the expression for conservation of energy to solve this problem:

Assuming a final height of zero, we can eliminate final potential energy. Then, substituting in expressions for each variable, we get:

$mgh_i+\frac{1}{2}mv_i^2=\frac{1}{2}mv_f^2$

Canceling out mass and rearranging for final velocity, we get:

$v_f =\sqrt{2gh_i+v_i^2} = \sqrt{2(10\frac{m}{s^2})(3m)+(8\frac{m}{s})^2}$

$v_f = 11.1\frac{m}{s}$

A man weighing walks to the top of a tall building. How much gravitational potential energy does he have at the top of the building compared to when he was on the street below?

Explanation

For this question we will use conservation of gravitational potential energy.

$P_{gravity}=mgh$

Plug in and solve for $P_{gravity}$ .

$P_{gravity}=90Kg(9.8\frac{m}{s^{2}})(527m)$

$P_{gravity}=464,814j$

The man will gain of gravitational potential energy by walking to the top of the building.

Planet A is twice as massive as Planet B. Compared to a person standing on Planet A, a person standing on Planet B will have __________ potential energy and __________ mass.

lower . . . the same

lower . . . lower

the same . . . lower

the same . . . the same

Explanation

Potential energy is calculated using the following equation.

$U = \frac{-GMm}{r}$

Where is the potential energy, is the gravitational constant, is the mass of one object, is the mass of the other object, and is the distance between the two objects. The potential energy depends on the mass of the objects (in this case the person and the planet); therefore, the more massive planet will produce the higher potential energy.

The mass is the same regardless of the situation. Mass measures the amount of substance/elements/molecules inside a person or object's body. The composition of the person’s body won’t change (regardless of the mass of the planet); therefore, the mass will stay the same. Recall that the weight of the person, however, does change based on the planet the person is standing on. This is because the weight depends on the gravitational acceleration, which depends on the mass of the planet.

$G=6.67*10^{-11}\frac{N*m^2}{kg^2}$

Mass of moon: $7.3*10^{22}kg$