Question

A body of mass 10 kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is *



(a) 1 m s^–2 at an angle of tan-^1 (4 / 3) w.r.t. 6N force.

(b) 0.2 m s^–2 at an angle of tan-^1 (4 / 3) w.r.t. 6N force.

(c) 1 m s^–2 at an angle of tan-^1 (3 / 4) w.r.t. 6N force.

(d) 0.2 m s^–2 at an angle of tan-^1 (3 / 4) w.r.t. 8N force​

Answer 1

Here, m=10 kg

The resultant force acting on the body is

F=(98N)2+(6N)2=10N

Let the resultant force F makes an angle θ w.r.t. 8N force.

From figure, tanθ=8N6N=43

The resultant acceleration of the body is 

a=mF=10kg10N=1ms−2

The resulatnt acceleration is along the direction of the resulatnt force.

Hence, the resultant acceleration of the body is 1 ms−2 at an angle of tan−1(43) w.r.t. 8N force.