A small block of mass m = 225 g is released from rest at point circled A along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 27.5 cm as shown in the figure below.
(a) Calculate the gravitational potential energy of the block-Earth system when the block is at point circled A relative to point circled B.
(b) Calculate the kinetic energy of the block at point circled B.
J
(c) Calculate its speed at point circled B.
m/s
(d) Calculate its kinetic energy when the block is at point circled C.
J
(e) Calculate the potential energy when the block is at point circled C.
J
To solve this problem, we can use the principles of conservation of energy. The total mechanical energy of the system is conserved as the block moves from point A to point B to point C.
(a) To calculate the gravitational potential energy of the block-Earth system when the block is at point A relative to point B, we need to find the difference in height between the two points. Since the block is on the inside of a hemispherical bowl with radius R, the height difference is equal to the radius of the bowl (R).
Gravitational potential energy (U) is given by the formula: U = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
In this case, m = 225 g = 0.225 kg, g = 9.8 m/s^2, and h = R = 27.5 cm = 0.275 m.
Therefore, the gravitational potential energy at point A relative to point B is:
U = mgh = (0.225 kg)(9.8 m/s^2)(0.275 m) = 0.596 J
(b) The kinetic energy of the block at point B can be calculated using the conservation of mechanical energy. Since the system is frictionless, the total mechanical energy (E) is conserved and is the sum of the gravitational potential energy (U) and kinetic energy (K) at point A. Therefore, E = U + K.
We already calculated U at point A, so we can write:
E = 0.596 J + K
At point B, the block is at the highest point of its motion, so its kinetic energy is zero. Therefore, at point B, K = 0 J.
Substituting K = 0 J into the equation, we get:
E = 0.596 J + 0 J = 0.596 J
Therefore, the kinetic energy of the block at point B is 0 J.
(c) The speed of the block at point B can be calculated using the conservation of mechanical energy. Since the system is frictionless, the total mechanical energy (E) is conserved and is the sum of the gravitational potential energy (U) and kinetic energy (K).
At point B, as mentioned earlier, the kinetic energy is zero. Therefore, E = U = 0.596 J.
Using the equation for gravitational potential energy, we can solve for the velocity (v) at point B:
E = U = mgh = (0.225 kg)(9.8 m/s^2)(R)
0.596 J = (0.225 kg)(9.8 m/s^2)(0.275 m)
v = sqrt(2gh)
Therefore, the speed of the block at point B is:
v = sqrt(2gh) = sqrt(2(9.8 m/s^2)(0.275 m)) ≈ 1.62 m/s
(d) To calculate the kinetic energy of the block when it is at point C, we again use the conservation of mechanical energy. The total mechanical energy (E) is still conserved and is equal to the sum of the gravitational potential energy (U) and kinetic energy (K) at point C.
Since the block is at the lowest point of its motion at C, its potential energy is zero (U = 0 J). Therefore, at point C, E = K.
Using the equation for the total mechanical energy, we can solve for the kinetic energy (K) at point C:
E = K = (0.596 J + 0 J) = 0.596 J
Therefore, the kinetic energy of the block at point C is 0.596 J.
(e) To calculate the potential energy when the block is at point C, we need to find the difference in height between C and the reference point B.
Using the same formula for gravitational potential energy as before (U = mgh), and since the height difference is the same as the radius of the bowl (R):
U = mgh = (0.225 kg)(9.8 m/s^2)(0.275 m) = 0.596 J
Therefore, the potential energy of the block when it is at point C is 0.596 J.