A wooden block of mass 8 kg slides down an inclined plane of inclination 30° to the horizontal with constant acceleration 0.4 m/s2. The force of friction between the block and inclined plane is (g = 10 m/s2)

Question

A wooden block of mass 8 kg slides down an inclined plane of inclination 30° to the horizontal with constant acceleration 0.4 m/s2. The force of friction between the block and inclined plane is (g = 10 m/s2) (A) 12.2 N (B) 24.4 N (C) 36.8 N (D) 48.8 N

Tardigrade Admin · Accepted Answer

Here, m g sin θ-f=m a <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fe76d29906b41c37fc02dd5468f2fe04-.png" /> m g sin θ-m a=f 8 × 10 sin (30°)-8 × 0.4=f 40-3.2=f ⇒ f=36.8 N

Q. A wooden block of mass $8 k g$ slides down an inclined plane of inclination $3 0^{\circ}$ to the horizontal with constant acceleration $0.4 m / s^{2}$ . The force of friction between the block and inclined plane is $(g = 10 m / s^{2})$

$12.2 N$

$24.4 N$

$36.8 N$

$48.8 N$

Here, $m g sin θ - f = ma$

$m g sin θ - ma = f$
$8 \times 10 sin (3 0^{\circ}) - 8 \times 0.4 = f$
$40 - 3.2 = f \Rightarrow f = 36.8 N$

Q. A wooden block of mass 8kg slides down an inclined plane of inclination 30∘ to the horizontal with constant acceleration 0.4m/s2. The force of friction between the block and inclined plane is (g=10m/s2)

12.2N

24.4N

36.8N

48.8N