A body of mass 10 kg is initially stationary on a 45° inclined plane as shown in figure. The coefficient of dynamic friction between the body and the plane is 0.3. The body slides down the plane and attains a velocity of 20 m/s. The distance travelled (in metre) by the body along the plane is close to

A body of mass 10 kg is initially stationary on a 45° inclined plane as shown in figure. The coefficient of dynamic friction between the body and the plane is 0.3. The body slides down the plane and attains a velocity of 20 m/s. The distance travelled (in metre) by the body along the plane is close to Correct Answer 41

Concept:

The FBD of diagram is shown below.

[ alt="Capture96" src="//storage.googleapis.com/tb-img/production/17/05/Capture96.PNG" style="height:262px; width:233px">

The body initially at rest position, so, u = 0

Applying force balance

N = mgcos θ and f = mg sinθ 

Now, mg sinθ – μN = ma................(1)

The velocity of body is, V2 = u2 + 2aS

Calculation:

Given:

m = 10 kg, v = 20 m/s, μ = 0.3, θ = 45° 

mg sinθ – μ mg cos θ = ma................(from 1)

⇒ a = g(sin θ - μ cos θ)

= 9.81 (sin45 – 0.3 cos 45)

a = 4.85 m/s²

The distance covered by body 

V² = u² + 2aS

⇒ 20² = 0 + 2 × 4.85 × S

⇒ S = 41.23 m.