Posted by **Lissa.** on Saturday, May 31, 2014 at 4:34pm.

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.

- Physics. -
**MathMate**, Saturday, May 31, 2014 at 6:35pm
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_{k}=0.25

Frictional force = μ_{k}m1g

Let tension in string, T

Acceleration, a = (T-μ_{k}mg)/m

=T/m1-μ_{k}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_{k}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_{k}=0.25

You should get acceleration, a=2.1 m/s² approximately.

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