First slide

Friction as component of Net Contact Force

Question

A force F equals to 100 N is applied on 30 kg box inside which a 10 kg mass is hanging with the help of a massless string as shown in figure. Coefficient of friction between box and surface is 0.1. Initially system is at rest, then at this instant acceleration of box will be

Moderate

A

4 m/s²
B

6 m/s²
C

2 m/s²
D

1.5 m/s ²

Solution

$\begin{array}{l} F_{net} = F - f = 100 - 40 = 60 N \\ \Rightarrow a = \frac{F_{net}}{m} = 2 m / s^{2} \end{array}$

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