In the following figure m1 = 20.0 and angle = 50.7 degrees. The coefficient of k

Question

In the following figure m1 = 20.0 and angle = 50.7 degrees. The coefficient of kinetic friction between the block and the incline is = 0.40. What must be the mass m2 of the hanging block if it is to descend 14.5 m in the first 3.00 safter the system is released from rest?

In the following figure m1 = 20.0 and angle = 50.7 degrees. The coefficient of kinetic friction between the block and the incline is = 0.40. What must be the mass m2 of the hanging block if it is to descend 14.5 m in the first 3.00 safter the system is released from rest?

Explanation / Answer

Find the acceleration of the system:

y = v+ 0.5at²--------------------v = 0, starts from rest
a = 2y/t²
= 2(12.0m)/(3.00s)²
= 2.67m/s²

Now, Newton’s 2nd law for the mass on the incline (parallel forces):

F(x) = ma = T - mgsin - n------------->eqn. 1a

For m perpendicular to the incline:

F(y) = 0 = n - mgcos
n = mgcos---------------------eqn.1b

Plugging 1b into 1a, we get:

ma = T - mgsin - mgcos----------------------------->eq…

For the hanging mass:

F = ma = mg - T----------------------->eqn.3

Adding eqn. 2 and 3 together eliminates T, so we can solve for m:
ma + ma = mg - mgsin - mgcos
ma - mg = -mg(sin + cos)
m = -mg(sin + cos) / (a - g)
= -(20.0kg)(9.81m/s²)(sin50.7° + 0.400cos50.7°) / (2.67m/s² - 9.81m/s²)
= 28.2kg

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