A block of mass M rests on a block of mass M1 = 5.00 kg

which is on a tabletop. A light string passes over
a frictionless peg and connects the blocks. The coefficient of
kinetic friction mk at both surfaces equals 0.330. A force of
60.0 N pulls the upper block to the left and the lower block to
the right. The blocks are moving at a constant speed. Determine
the mass of the upper block.

I'm trying to combine the equations
f(k) = mu(k) * F(N)
and
F(N) = ( M(1) + M ) * mu * g
to solve for M, but I keep getting really large numbers. :/

Jake how did you get that answer?

3.73

To solve the problem, we can start by analyzing the forces acting on each block separately.

For the upper block (mass M), the force is 60.0 N to the left, and the force of kinetic friction is pointing to the right. Since the blocks are moving at a constant speed, the forces must be balanced. Therefore, we have:

Force to the left = Force of kinetic friction to the right

60.0 N = μk * FN

Where FN is the normal force acting on the upper block.

Now, for the lower block (mass M1 = 5.00 kg), the force is 60.0 N to the right, and the force of kinetic friction is pointing to the left. Again, since the blocks are moving at a constant speed, these forces must be balanced. Therefore, we have:

Force to the right = Force of kinetic friction to the left

60.0 N = μk * FN1

Where FN1 is the normal force acting on the lower block.

To calculate the normal forces, we can use the equation:

FN = (M + M1) * g

Now we can solve for the masses of the blocks.

Step 1: Calculate the normal forces.
FN = (M + M1) * g
FN = (M + 5.00 kg) * 9.8 m/s^2

Step 2: Substitute the normal forces into the equations for kinetic friction.
60.0 N = μk * [(M + 5.00 kg) * 9.8 m/s^2]
60.0 N = μk * (M * 9.8 m/s^2 + 5.00 kg * 9.8 m/s^2)
60.0 N = μk * (M * 9.8 m/s^2 + 49.0 N)

Step 3: Solve for M.
60.0 N = μk * (M * 9.8 m/s^2 + 49.0 N)
60.0 N = 0.330 * (M * 9.8 m/s^2 + 49.0 N)
60.0 N = 0.330M * 9.8 m/s^2 + 0.330 * 49.0 N
60.0 N - 0.330 * 49.0 N = 0.330M * 9.8 m/s^2
60.0 N - 16.17 N = 3.234M
43.83 N = 3.234M

M = 43.83 N / 3.234

M ≈ 13.535 kg

Therefore, the mass of the upper block is approximately 13.535 kg.

To solve this problem, you are on the right track of combining the equations. Let's break down the steps to find the mass of the upper block.

First, we need to calculate the normal force acting on the upper block. The normal force is equal to the weight of the upper block, which is given by:

F(N) = M * g

where M is the mass of the upper block and g is the acceleration due to gravity.

Next, we can use the equation for kinetic friction to find the frictional force acting on the upper block:

f(k) = μ(k) * F(N)

where f(k) is the frictional force and μ(k) is the coefficient of kinetic friction.

Since the lower block is being pulled with a force of 60.0 N to the right, the upper block experiences an equal and opposite force of 60.0 N to the left.

So, we have:

f(k) = 60.0 N

Now, substituting the expressions for F(N) and f(k) into the equation for kinetic friction:

μ(k) * M * g = f(k) = 60.0 N

Rearranging the equation, we get:

M * g = f(k) / μ(k)

Finally, we can solve for the mass of the upper block, M:

M = (f(k) / μ(k)) / g

Substituting the given values of f(k) = 60.0 N, μ(k) = 0.330, and the acceleration due to gravity g = 9.8 m/s², we can calculate the mass of the upper block:

M = (60.0 N / 0.330) / 9.8 m/s²

M ≈ 1.82 kg

Therefore, the mass of the upper block is approximately 1.82 kg.