question_answer

A force of 100 N is applied on a block of mass 3 kg as shown in the figure. The coefficient of friction between the wall and block is\[1/4\] .The frictional force acting on the block is

A)  20 N downwards       

B)  25 N upwards

C)  15 N downwards       

D)  20 N upwards

Correct Answer: A

Solution :

As shown in the figure. Normal reaction \[N=F\cos {{30}^{o}}=\frac{100\sqrt{3}}{2}\text{newton}\] \[{{f}_{\max }}=\mu N=\frac{1}{4}\times \frac{100\sqrt{3}}{2}=\frac{25}{2}\sqrt{3}\,N\] Now, net force acting on the block excluding friction \[=F\sin {{30}^{o}}-3g\] \[=100\times \frac{1}{2}-3\times 10=20\,N,\]upwords Hence, the correction option is [a].