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 inclined plane? [g=10 ms -2]

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 inclined plane? [g=10 ms -2] (A) 36.8 N (B) 76.8 N (C) 65.6 N (D) 97.8 N

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/4c2873472cf39bf2f174ee5a3e047fde-.png" /> m g sin θ-μ N=m a ⇒ μ N=m g sin θ-m a =(8 kg )(10 ( m / s 2)) sin 30°-(8 kg )(0.4 ms -2)=40 N -3.2 N ⇒ Frictional force =36.8 N

Q. A wooden box of mass $8 k g$ slides down an inclined plane of inclination $3 0^{\circ}$ to the horizontal with a constant acceleration of $0.4 m s^{- 2} .$ What is the force of friction between the box and inclined plane? $[g = 10 m s^{- 2}]$

36.8 N

76.8 N

65.6 N

97.8 N

$m g sin θ - μ N = ma$
$\Rightarrow μ N = m g sin θ - ma$
$= (8 k g) (10 \frac{m}{s ^{2}}) sin 3 0^{\circ} - (8 k g) (0.4 m s^{- 2}) = 40 N - 3.2 N$
$\Rightarrow$ Frictional force $= 36.8 N$

Q. A wooden box of mass 8kg slides down an inclined plane of inclination 30∘ to the horizontal with a constant acceleration of 0.4ms−2. What is the force of friction between the box and inclined plane? [g=10ms−2]