At position A, the ball has only potential energy due to its position above the ground.
The potential energy of the ball at position A is:
PE = mgh
where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the ball above the ground.
PE(A) = 8 kg * 9.8 m/s^2 * 5 m
PE(A) = 392 J
At position B, the ball has both kinetic and potential energy.
The potential energy of the ball at position B is:
PE = mgh
where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the ball above the ground.
PE(B) = 8 kg * 9.8 m/s^2 * 2.5 m
PE(B) = 196 J
The kinetic energy of the ball at position B is:
KE = (1/2)mv^2
where m is the mass of the ball and v is the velocity of the ball.
Since the ball is rolling down the ramp, we can use the conservation of energy principle to find the velocity of the ball at position B.
The total energy of the ball at position A is:
E(A) = PE(A)
E(A) = 392 J
The total energy of the ball at position B is:
E(B) = PE(B) + KE(B)
E(B) = 196 J + (1/2)(8 kg)v^2
E(B) = 196 J + 4v^2
Since energy is conserved, we have:
E(A) = E(B)
392 J = 196 J + 4v^2
196 J = 4v^2
v^2 = 49 m^2/s^2
v = 7 m/s
The kinetic energy of the ball at position B is:
KE(B) = (1/2)mv^2
KE(B) = (1/2)(8 kg)(7 m/s)^2
KE(B) = 196 J
At position C, the ball has only kinetic energy.
The potential energy of the ball at position C is zero since the ball is at ground level.
The kinetic energy of the ball at position C is:
KE(C) = (1/2)mv^2
KE(C