Consider the system shown in the figure with m1 = 25.00 kg, m2 = 10.20 kg, R = 0.18 m, and the mass of the uniform pulley M = 5.00 kg. Object m2 is resting on the floor, and object m1 is 4.00 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) Find the time interval for m1 to hit the ground
b) Find the time interval if the pulley were massless.
Please help me out. This is the only problem I REALLY DON'T GET!!
The system is an Atwood machine, by the way.
To find the time interval for object m1 to hit the ground, we can use the concept of conservation of mechanical energy.
First, let's consider the forces acting on the system. The weight of m1 (mg1), m2 (mg2), and M (Mg) are acting downward. The tension in the rope (T) is acting upward on m1 and downward on m2.
a) Find the time interval for m1 to hit the ground:
1. We start by calculating the acceleration of the system using Newton's second law. The net force on the system is given by the difference in the weight of m1 and m2:
F_net = m1 * g - m2 * g
where g is the acceleration due to gravity (9.8 m/s^2) and m1 and m2 are the masses of objects m1 and m2, respectively.
Re-arranging the equation, we get:
F_net = (m1 - m2) * g
2. Use this net force to find the total acceleration of the system by dividing the net force by the total mass of the system:
a = F_net / (m1 + m2 + M)
3. Next, we can calculate the initial potential energy of m1 using the formula:
PE_initial_m1 = m1 * g * h
where h is the initial height of m1 (4.00 m).
4. Then, calculate the initial kinetic energy of the system, which is 0 since the system is at rest initially (KE_initial = 0).
5. Since energy is conserved, the initial mechanical energy (E_initial) is equal to the sum of the initial potential energy and initial kinetic energy:
E_initial = PE_initial_m1 + KE_initial
6. At the point where m1 hits the ground, its height is zero. Therefore, the final potential energy of m1 is zero (PE_final_m1 = 0).
7. We can now use the principle of conservation of mechanical energy to equate the initial and final mechanical energies:
E_initial = PE_final_m1 + KE_final_m1
8. Substitute the expressions for initial and final energies:
PE_initial_m1 + KE_initial = PE_final_m1 + KE_final_m1
9. Rearrange the equation to solve for final kinetic energy (KE_final_m1):
KE_final_m1 = PE_initial_m1 - PE_final_m1
10. The final kinetic energy (KE_final_m1) can be calculated using:
KE_final_m1 = 0.5 * m1 * v^2
where v is the final velocity of m1 just before it hits the ground.
11. Substitute the expressions for KE_final_m1 and PE_final_m1 derived in steps 9 and 6 respectively:
0.5 * m1 * v^2 = PE_initial_m1
12. Solve for v^2:
v^2 = 2 * (PE_initial_m1 / m1)
13. Take the square root of both sides to find the final velocity (v):
v = √(2 * (PE_initial_m1 / m1))
14. Now, we can calculate the time interval for m1 to hit the ground using the formula for time:
t = d / v
where d is the distance traveled by m1 (which is equal to the initial height h) and v is the final velocity obtained in step 13.
Substituting the values given (m1 = 25.00 kg, m2 = 10.20 kg, R = 0.18 m, M = 5.00 kg, and h = 4.00 m) should allow you to find the final answer for part a).