Assuming that the string connecting the two blocks does not stretch, we can apply the principle of conservation of mechanical energy to find the speed of the blocks after sliding a distance of 0.25 m.
The potential energy of block A at the starting position is:
U1 = m_A * g * 0 = 0
where g is the acceleration due to gravity (9.8 m/s^2).
When block A slides a distance s = 0.25 m down the incline, the potential energy is converted into kinetic energy:
K1 = U1 - W_friction
where W_friction is the work done by friction (which is zero in this case).
The kinetic energy of block A at the bottom of the incline is:
K2 = (1/2) * m_A * v_A^2
where v_A is the speed of block A.
The potential energy of block B at the starting position is:
U1 = m_B * g * 0 = 0
When block A slides a distance s = 0.25 m down the incline, the potential energy of block B increases by the same amount:
U2 = m_B * g * s = 245 J
The kinetic energy of block B at the bottom of the incline is:
K2 = (1/2) * m_B * v_B^2
where v_B is the speed of block B.
Since the two blocks are connected by a string, their speeds must be the same:
v_A = v_B = v
Conservation of mechanical energy states that the total initial mechanical energy (potential energy of block A at the starting position) is equal to the total final mechanical energy (kinetic energy of both blocks at the bottom of the incline):
U1 = K2 = (1/2) * m_A * v^2 + (1/2) * m_B * v^2 = (1/2) * (m_A + m_B) * v^2
Substituting the given values, we have:
0 = (1/2) * 50 * v^2 + (1/2) * 100 * v^2 - 245
Simplifying:
75 * v^2 = 245
v^2 = 245/75
v = sqrt(3.27) = 1.81 m/s
Therefore, the speeds of both blocks after A has slid a distance of 0.25 m are 1.81 m/s.
Two
blocks, A and B (m A 50 kg and m B 100 kg), are
connected by a string as shown If the blocks begin at rest,
what will their speeds be after A has slid a distance
𝑠=0.25𝑚 Assume the pulley and incline are frictionless
1 answer
1 answer