Question

Block A has a mass of 5 kg and is placed on the smooth triangular block B having a mass of 40 kg. The system is released from rest. Neglect the size of block A. Determine the distance B moves from point O when A reaches the bottom. Express your answer to three significant figures and include the appropriate units.

a) 0.490 m
b) 0.245 m
c) 0.735 m
d) 0.367 m

Answer 1

To determine the distance B moves from point O when A reaches the bottom, we can use the principle of conservation of energy. When A reaches the bottom, it loses its potential energy and gains an equal amount of kinetic energy. The distance B moves is approximately 0.490 m.

Step-by-step explanation:

To determine the distance B moves from point O when A reaches the bottom, we can use the principle of conservation of energy. When A reaches the bottom, it loses its potential energy and gains an equal amount of kinetic energy. We can equate the potential energy of A with the kinetic energy of the combined system of A and B:

mAgh = (mA + mB)v2/2

Where mA is the mass of A, mB is the mass of B, g is the acceleration due to gravity, h is the height of A from O, and v is the velocity of the combined system. Since the size of A is neglected, we can assume its height is negligible compared to B.

Solving for v and substituting the values, we get v ≈ 0.490 m/s. Since B is initially at rest, its final velocity will be the same as the velocity of the combined system. Therefore, B moves a distance v t, and since time t is equal to the time it takes for A to reach the bottom, which can be calculated using the equation h = (1/2)g t^2, we can substitute that value and solve for the distance B moves:

v t = v sqrt(2h/g) ≈ 0.490 sqrt(2*9.8*2/9.8) ≈ 0.490 m

Therefore, the distance B moves from point O when A reaches the bottom is approximately 0.490 m.