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
46.0 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.
To determine the mass of the upper block (M), we can use the concept of equilibrium.
1. Start by drawing a free-body diagram for each block separately.
- The upper block (M) experiences a force of 46.0 N to the left and kinetic friction (fk) to the right.
- The lower block (M1) experiences kinetic friction (fk) to the left.
2. Write the equations of equilibrium for each block.
For the upper block (M):
46.0 N - fk = 0 (Equation 1)
For the lower block (M1):
fk - fk = 0 (Equation 2)
3. Solve Equation 1 for fk:
fk = 46.0 N
4. Substitute the value of fk from Equation 1 into Equation 2:
46.0 N - fk = 0
5. Solve for fk:
46.0 N - fk = 0
fk = 46.0 N
6. Substitute the value of fk into Equation 1:
46.0 N - 46.0 N = 0
0 = 0
7. Since both equations are satisfied, it means that the system is in equilibrium.
8. From the equations, we can see that the force applied (46.0 N) is equal to the force of kinetic friction (fk).
9. The force of kinetic friction can be calculated using the formula:
fk = μk * N
where μk is the coefficient of kinetic friction and N is the normal force.
10. The normal force can be calculated as the weight of the lower block (M1). The weight (Fg) is given by:
Fg = m1 * g
where m1 is the mass of the lower block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
11. Substitute the known values into the equation:
Fg = 5.00 kg * 9.8 m/s^2
Fg = 49.0 N
12. Substitute the value of Fg into the formula for kinetic friction:
fk = μk * N
46.0 N = 0.330 * 49.0 N
13. Solve for the mass of the upper block (M):
M = fk / g
M = 46.0 N / 9.8 m/s^2
M = 4.69 kg
Therefore, the mass of the upper block is approximately 4.69 kg.