Question

A wooden box of mass 8 kg slides down an inclined plane of inclination 30 degree to the horizontal with a constant acceleration of 0.4 m per second square what is the force of friction between the box and the inclined plane

Answer 1

Answer:F=ma mgsin30° - Fric = ma 8×10×1/2 - Fric = 8×0.4 40 - Fric = 3.2 40 - 3.2 = Fric Friction = 36.8

Answer 2

36.8 N is the force of friction between the box and the inclined plane.

Explanation:

Given that,

m = 8kg

Inclination; ∅ = 30°

Acceleration; a = 0.4 m/$s^{2}$

g = 10 m/$s^{2}$

As we know,

m = mg sin ∅ - $F_{f}$

m = mg sin 30°  - $F_{f}$

8 * 0.4 = (8)(10) * 1/2  - $F_{f}$

 $F_{f}$ = 40 - 3.2

Therefore, the force of friction between  the box and the inclined plane:

$F_{f}$ = 36.8 N

Learn more: Frictional Force

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