In order for the mass M to descend at a constant velocity when released, the forces acting on it must be balanced. This means that the force of tension in the rope connecting mass M to the other blocks is equal to the gravitational force acting on mass M.
The total mass of the system that causes the tension is the sum of the 9 kg and 12 kg blocks, which is 21 kg. Now, the force of tension required to keep the system moving at a constant velocity comes from the friction between the 12 kg block and the surface.
The frictional force can be calculated using the formula: friction = μ * normal force, in this case the normal force is the force from the weight of the 12 kg block, which can be calculated by multiplying the mass (12 kg) by the gravitational acceleration (9.81 m/s²).
Normal force = 12 kg * 9.81 m/s² = 117.72 N
Friction = frictional coefficient * normal force = 0.30 * 117.72 N = 35.316 N
Since the force of tension is equal to the frictional force, we can now solve for the mass M.
Tension = M * g
35.316 N = M * 9.81 m/s²
M = 35.316 N / 9.81 m/s²
M ≈ 3.6 kg
The mass M that descends at a constant velocity when released is closest to 3.6 kg.
A system comprising blocks, a light frictionless pulley, and connecting ropes is shown. The 9 kg block is on a smooth horizontal table The surfaces of the 12 kg block are rough, with μ = 0.30.
In Figure 5.7, the mass M is set so that it descends at constant velocity when released. The mass M is closest to:
1 answer