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](https://tex.z-dn.net/?f=PE%3Dmgh)
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](https://tex.z-dn.net/?f=g%3D9.8%20m%2Fs%5E2)
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](https://tex.z-dn.net/?f=PE%3D%280.5%29%289.8%29%281%29%3D4.9%20J)
b)
The mechanical energy of an object is given by the sum of potential energy (PE) and kinetic energy (KE):
![E=PE+KE](https://tex.z-dn.net/?f=E%3DPE%2BKE)
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.](https://tex.z-dn.net/?f=E%3DPE%2BKE%3Dconst.)
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}](https://tex.z-dn.net/?f=E%3DPE_%7Btop%7D)
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}](https://tex.z-dn.net/?f=E%3DKE_%7Bbottom%7D)
This means that
![KE_{bottom}=PE_{top}](https://tex.z-dn.net/?f=KE_%7Bbottom%7D%3DPE_%7Btop%7D)
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](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
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](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B2KE%7D%7Bm%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%284.9%29%7D%7B0.5%7D%7D%3D4.4%20m%2Fs)
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](https://tex.z-dn.net/?f=PE%27%3Dmgh%27%3D%280.5%29%289.8%29%280.8%29%3D3.9%20J)
This also means that the total mechanical energy at the top (h'=0.8 m) is
![E'=3.9 J](https://tex.z-dn.net/?f=E%27%3D3.9%20J)
However, we are told that the kinetic energy of the ball when it is released is
![KE=4.9 J](https://tex.z-dn.net/?f=KE%3D4.9%20J)
So the mechanical energy at the release is
![E=4.9 J](https://tex.z-dn.net/?f=E%3D4.9%20J)
Therefore, the energy lost to friction is equal to the difference in energy:
![\Delta E=E-E'=4.9-3.9=1.0 J](https://tex.z-dn.net/?f=%5CDelta%20E%3DE-E%27%3D4.9-3.9%3D1.0%20J)
Learn more about potential and kinetic energy:
brainly.com/question/1198647
brainly.com/question/10770261
brainly.com/question/6536722
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