JEE Main Physics question
A wooden box of mass 8 kg slides down an inclined plane of inclination 30° to the horizontal with a constant acceleration of 0.4 ms⁻². What is the force of friction between the box and the inclined plane? (g = 10 ms⁻²)
Answer: (A)
Solution
mg sinθ − f = ma ⟹ f = 8×10×0.5 − 8×0.4 = 40 − 3.2 = 36.8 N.