A block of mass 10kg is pressed against a vertical wall. If the coefficient of friction between the wall and the block is 0.2, then what is the minimum force that should be applied on the block so that does not fall to the ground?
A
500N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
400N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
600N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
555.5N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A500N By the FBD of the block we get,
The force F is applied against the wall, which produces a normal reaction N. Frictional force arises due to this reaction force and counters the weight of the block.
Hence for equillibrium,
N=F μN=mg μF=mg⇒F=mgμ=1000.2=500N
Hence a force of 500N has to be applied to keep the block in equilibrium.