Consider the mass and pulley system shown. Mass m 1 = 25kg and mass m 2 = 14kg .

Question

Consider the mass and pulley system shown. Massm1 = 25kg and mass m2 = 14kg . The angle of the inclined plane is given, and the coefficient of kinetic friction between mass m2 and the inclined plane is ?k=0.19. Assume the pulleys are massless and frictionless.

What is the acceleration of mass m2 on the inclined plane? Take positive acceleration to be up the ramp.

If the system is released from rest, what is the speed of mass m2 after 3s ?

When mass m2 moves a distance 4.18m up the ramp, how far downward does mass m1 move?

Explanation / Answer

Let T = tension in string.
FBD of m1 yields, m1*g - 2T = m1*a
Putting values, 25*9.81 - 2T = 25*a.............1

Normal force on m2 = m2*g*cos theta
Friction force = 0.19*normal force = 0.19*m2*g*cos theta

FBD of m2 yields, T - m2*g*sin theta - 0.19*m2*g*cos theta = m2*a

Putting values, T - 14*9.81*(5/13) - 0.19*14*9.81*(12/13) = 14*a..................2

Solving eqns 1 and 2 we get,
a = 1.725 m/s^2
T = 101.1 N

After 3 s, speed of mass m2 is = a*t = 1.725*3 = 5.175 m/s

Time to cover 4.18 m is, 4.18 = 1/2*at^2
4.18 = 1/2*1.725*t^2
t = 2.2 s

In 2.2 s, height dropped by m1 is given by 2.2*1.725 = 3.797 m

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