Question

A 5.38 kg block is placed on top of a 9.24 kg block. A horizontal force of F = 70.6 N is applied to the 9.24 kg block and the 5.38 kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is 0.28. There is friction both between the masses and between the 9.24 kg block and the ground. The gravity acceleration is 9.8 m/s^2. Determine the magnitude of the acceleration of the 9.24 kg block. 1.7 m/s^2 2.6 m/s^2 4.9 m/s^2 7.4 m/s^2 None of the above

Answer 1

Horizontal force $=F=70.6N$

Coefficient of kinetic friction $=\mu =0.28$

Acceleration due to gravity $=g=9.8m/s^{2}$

Total frictional force on 9.24 kg block $=f_{s}$

$f_{s}=\left [ 0.28\times (5.38+9.24)\times 9.8+5.38\times 9.8\times 0.28 \right ]N=54.88N$

Resultant force on 9.24kg block $=F_{r}$

$F_{r}=F-f_{s}=(70.6-54.88)N=15.72N$

Acceleration on that block $=a$

$a=\frac{15.72}{9.24}m/s^{2}=1.7m/s^{2}$

Hence a. is the right answer