Question

The figure shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angle ? = 20

Answer 1

In order for the block to move, one basic condition must be fulfilled i.e. the force applied in the direction of motion should be greater than the maximum static friction that acts on the box.

The force that is applied on the box can be divided into two components

The vertical, F_V = 0.5mgsin20 = 0.17mg

The horizontal, F_H = 0.5mgcos20 = 0.47mg

We have two force relations on the block

In the vertical direction,

N + F_V = mg

N + 0.17mg = mg

N = 0.83mg

The maximum static friction is a product of the normal reaction on the block and coefficient of static friction.

If the horizontal force acting on the block is more than the max. static friction, then the block moves and dynamic friction acts on it, in which case the frictional force acting on the block is a product of the dynamic friction coefficient and normal reaction on the block. Else, the block does not move and acceleration is zero.

a)

So, F_H = 0.47mg < f_max

Hence, the block does not move and the magnitude of acceleration is zero.

Ans. a = 0.

b)

F_H = 0.47mg > f_max

Hence, the block moves.

Calculating the value of dynamic friction, we get

Equating the forces in the horizontal direction, we get

F_H - f = ma

0.47mg - 0.27mg = ma

0.2mg = ma

a = 0.2g

At g = 9.8 m/s²

a = 0.2 x 9.8 = 1.96 m/s²

Ans. a = 1.96 m/s²

Equating the forces in