Two Body, Friction (10 points)?

This is a messy problem, but here is a solution. It may not be the simplest but it gives the correct answer.

There are 2 horizontal forces - the 16N forwards and kinetic frictional force backwards.

Call the coefficient of kinetic friction μ and take g=9.81m/s².

If the sled has mass m then its weight is mg = 9.81m.

Since there is no vertical acceleration, the normal force is 9.81m.

The kinetic frictional force is therefore 9.81mμ

So the resultant force causing acceleration is 16 - 9.81mμ

16 - 9.81mμ = ma

16 - 9.81mμ = 0.40m

9.81mμ = 16 - 0.40m (equation 1)

With the extra mass of 4.2kg, the new mass is m+4.2.

The new weight is = 9.81(m+4.2)

The new normal force is 9.81(m+4.2)

The new kinetic frictional force is therefore 9.81(m+4.2)μ

The new resultant force is 16 - 9.81(m+4.2)μ

16 - 9.81(m+4.2)μ = ma

There is zero acceleration so

16 - 9.81(m+4.2)μ = 0

9.81(m+4.2)μ = 16 (equation 2)

The algebra gets messy. Here's one method:

If we subtract equation 1 from equation 2 we get

[ 9.81(m+4.2)μ] - [9.81mμ] = [16] - [(16-0.40m)]

41.2μ = 0.40m

m = 103μ

Substituting for m in equation 1

9.81(103μ) μ = 16-0.40(103μ)

1010μ² = 16 - 41.2μ

1010μ² + 41.2μ - 16 = 0

Solving the quadratic equation in the usual way gives

μ = 0.107 or -0.148.

We ignore the negative solution as it is not physically meaningful, so the answer is 0.107.

let m = mass of sled

F = ma <=> 16 = m(0.40)

m = 16/0.40 = 40 kg

Normal force of sled & extra mass = 44.2 (9.81) = 434 N

applied horizontal force = 16 N

friction force = 16 N because applied force barely moves sled

(proper physics speak would be to say same force moves sled at "constant speed"}

µk = friction force/Normal force = 16/434 = 0.037 ANS

Sorry I can't explain how to get 0.107 for the answer :>)