Question

A very thoughtful physics student takes her younger sibling to the arcade to play skeeball. She estimates the mass of the skee ball to be about 0.5 kg and the height of the fifty-point ring to be about 1 m above the position from which the ball is released.
a. How much gravitational potential energy will the ball have if it hits the fifty-point ring at the highest point?

b. How much kinetic energy does the ball need at its release to reach the highest point?

c. At what velocity must the ball be released to reach the highest point?

d. The student’s sibling uses a radar gun to determine how fast she releases the ball, and it is exactly the speed she calculated to reach the 1 m height. However, the ball lands in the thirty-point ring, which is 0.8 m above the release point. How much energy is lost to friction if the ball only reaches 0.8 m instead of 1 m?

Answer 1

a) Gravitational potential energy: 4.9 J

b) Kinetic energy: 4.9 J

c) Velocity of the ball: 4.4 m/s

d) Energy lost to friction: 1.0 J

Explanation:

a)

The gravitational potential energy of an object is the energy possessed by the object due to its position in the gravitational field. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

For the ball in this problem,

m = 0.5 kg

$g=9.8 m/s^2$

h = 1 m (the ball is at the 50-point ring)

Therefore, the potential energy is

$PE=(0.5)(9.8)(1)=4.9 J$

b)

The mechanical energy of an object is given by the sum of potential energy (PE) and kinetic energy (KE):

$E=PE+KE$

where the kinetic energy is the energy due to the motion.

In absence of frictional force, the total mechanical energy of the ball is constant:

$E=PE+KE=const.$

At the maximum height, the kinetic energy is zero (since the ball changes direction), so all the mechanical energy is potential energy:

$E=PE_{top}$

However, when the ball is released, h = 0, so the potential energy is zero and all the mechanical energy is kinetic energy:

$E=KE_{bottom}$

This means that

$KE_{bottom}=PE_{top}$

Therefore, the kinetic energy of the ball at its release is  also 4.9 J.

c)

The kinetic energy of an object is given by

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

where

m is the mass of the object

v is its speed

In this problem, we have

m = 0.5 kg is the mass of the ball

KE = 4.9 J is the kinetic energy

Therefore, the velocity of the ball at its release is

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(4.9)}{0.5}}=4.4 m/s$

d)

If the ball reaches only a height of h = 0.8 m, then its potential energy at the top is

$PE'=mgh'=(0.5)(9.8)(0.8)=3.9 J$

This also means that the total mechanical energy at the top (h'=0.8 m) is

$E'=3.9 J$

However, we are told that the kinetic energy of the ball when it is released is

$KE=4.9 J$

So the mechanical energy at the release is

$E=4.9 J$

Therefore, the energy lost to friction is equal to the difference in energy:

$\Delta E=E-E'=4.9-3.9=1.0 J$

Learn more about potential and kinetic energy:

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