Consider the system shown in the figure below with m1 = 30.0 kg, m2 = 13.7 kg, R = 0.110 m, and the mass of the pulley M = 5.00 kg. Object m2 is resting on the floor, and object m1 is 4.60 m above the floor when it is released from rest. The pulley axis is frictionless. The cord is light, does not stretch, and does not slip on the pulley.
(a) Calculate the time interval required for m1 to hit the floor.
Δt1 = s
(b) How would your answer change if the pulley were massless?
Δt2 = s
To solve this problem, we can use the principles of conservation of energy and Newton's laws of motion. Let's break it down step by step.
(a) Calculate the time interval required for m1 to hit the floor:
To find the time it takes for m1 to hit the floor, we can use the principle of conservation of energy. At the top, m1 has potential energy given by m1*g*h (where g is the acceleration due to gravity and h is the height). At the bottom, all of this potential energy is converted into kinetic energy, given by (1/2)*m1*v^2 (where v is the velocity).
1. Find the velocity of m1 just before it hits the floor:
Using conservation of energy, we can equate the potential energy at the top to the kinetic energy at the bottom:
m1*g*h = (1/2)*m1*v^2
Rearrange the equation to solve for v:
v = sqrt(2*g*h)
Here, g = 9.8 m/s^2 (acceleration due to gravity) and h = 4.6 m (height).
Substitute the values to find v:
v = sqrt(2*9.8*4.6) = 9.8 m/s (approx)
2. Find the time it takes for m1 to hit the floor:
Using the kinematic equation for motion under constant acceleration, we can find the time taken (Δt):
Δt = v / g
Substitute the values to find Δt:
Δt = 9.8 / 9.8 = 1 s
Therefore, the time interval required for m1 to hit the floor is 1 second (s).
(b) How would your answer change if the pulley were massless:
If the pulley were massless, it means that the mass of the pulley (M = 5.00 kg) does not contribute to the system's dynamics. In this case, the only forces acting on the system are the tensions in the cord.
Since the mass of the pulley no longer affects the system, the tension in the cord remains the same throughout the motion. Therefore, the time interval required for m1 to hit the floor remains the same, which is 1 second (s).
Therefore, both Δt1 and Δt2 in this case would be 1 second (s).