Asked by Lissa.

A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest.

Answered by MathMate
Start with a free-body diagram (FBD).
It solves most statics problems automatically when you isolate the system into individual pieces.

Start with the 500g block, mass m1.
Normal force=m1g
coefficient of friction, μ<sub>k</sub>=0.25
Frictional force = μ<sub>k</sub>m1g
Let tension in string, T
Acceleration, a = (T-μ<sub>k</sub>mg)/m
=T/m1-μ<sub>k</sub>g)

For the block at the other end of the string.
Net vertical force
=m2g-T
Acceleration, a
=(m2g-T)/m2
=g-T/m2

Since two accelerations must be equal, we have
T/m1-μ<sub>k</sub>g) = g-T/m2

Solve for t and substitute into formulas above to get a, acceleration.
m1=0.500 kg
m2=0.300 kg
μ<sub>k</sub>=0.25

You should get acceleration, a=2.1 m/s² approximately.
Answered by bryan
2.14
Answered by WalnerMerine
If the coefficient of sliding friction between a 25 kg crate and the floor is 0.45, how much force is required to move the crate at a constant velocity across the floor?
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