block b is released from rest. after it drops 2.0 m...?
Block B is released from rest . After it drops 2.0 meters straight down, determine the velocity of Block A and Block B. Block A is on a horizontal surface with a pulley connecting the blocks. Assume the following values: mA=5kg, mB=8kg, uk=0.15.
- NCSLv 74 weeks ago
Assuming a massless pulley, the system mass is
M = (5 + 8) kg = 13 kg
The motive force is
F = mB*g = 8kg * 9.8m/s² = 78 N
and the resistive force is
f = µ*mA*g = 0.12 * 5kg * 9.8m/s² = 5.9 N
making the net force
Fnet = F - f = 73 N
so the system acceleration is
a = F / M = 5.6 m/s²
v² = u² + 2as = 0² + 2 * 5.6m/s² * 2.0m = 22 m²/s²
speed v = 4.7 m/s
Since velocity yas a direction, we can say that
A's velocity is 4.7 m/s toward the pulley
and B's velocity is 4.7 m/s down.
Hope this helps!