In the following figure m1 = 20.0 kg and a = 50.7 degrees. The coefficient of kinetic friction between the block and the incline is mu k = 0.40.
What must be the mass m2 of the hanging block if it is to descend 12.0 m in the first 3.00 s after the system is released from rest?
What must be the mass m2 of the hanging block if it is to descend 12.0 m in the first 3.00 s after the system is released from rest?
-
For block2 with mass m2:
3 forces: -mg, T2 and -ma
y = vt - ½at²
12 = 0 - ½a(3)²
Net acceleration = a = -2.67 m/s²
T = m2a = -2.67 * m2
For block1 with mass m1:
3 forces: Ff, m1gsinθ, T
T - Ff - m1gsinθ = 0
m2a - μm1gcosθ - m1gsinθ = 0
2.67 * m2 = (0.40)(20.0)(9.81)(cos 50.7º) + (20.0)(9.81)(sin 50.7º)
2.67 * m2 = 49.708 + 151.827
2.67 * m2 = 201.535
m2 = 75.48 kg
.
.
3 forces: -mg, T2 and -ma
y = vt - ½at²
12 = 0 - ½a(3)²
Net acceleration = a = -2.67 m/s²
T = m2a = -2.67 * m2
For block1 with mass m1:
3 forces: Ff, m1gsinθ, T
T - Ff - m1gsinθ = 0
m2a - μm1gcosθ - m1gsinθ = 0
2.67 * m2 = (0.40)(20.0)(9.81)(cos 50.7º) + (20.0)(9.81)(sin 50.7º)
2.67 * m2 = 49.708 + 151.827
2.67 * m2 = 201.535
m2 = 75.48 kg
.
.