Question

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 μk at both surfaces equals 0.330. A force of F = 56.0 N pulls the upper block to the left and the lower block to the right. The blocks are moving at a constant speed.
Required:
Determine the mass of the upper block.

Answer 1

Answer:

The mass of the upper block M = 4.12 kg

Explanation:

From the given information:

The upper block formula can be computed as :

$F = T + F_{f1}$

$F - T - F_{f1} =0$

where;

$F_{f1} = \mu Mg$

$F - T - \mu Mg=0 ---- (1)$

The lower block formula is as follows:

$T - F_{f1} -F_{f2} = 0$

$T - \mu Mg - \mu (M_1+M)g= 0$

$T = \mu Mg - \mu (M_1+M)g --- (2)$

Equating equation (1) and  (2) together, we have:

$F - \mu Mg - \mu Mg - \mu (M_1 + M) g = 0$

$56 - 3 Mg - \mu M_1g= 0$

$M = \dfrac{56 - \mu M_1 g}{3 \mu g}$

$M = \dfrac{56 - (0.330)(5)(9.8)}{3 (0.330) (9.8)}$

$M = \dfrac{56 - 16.17}{9.702}$

$M = \dfrac{39.83}{9.702}$

M = 4.12 kg