Question

A body with an initial speed of 10m / s is thrown upwards along an inclined plane of 30 and 20 m in length. What must be the coefficient of friction for the body to stop just at the end of the inclined plane?

A. 0.28

B. 0.18

C. 0.62

D. 0.51

E. 0.15

Answer 1

Answer : (A) 0.28

Solution :

Given :

u = 10 m/s

θ = 30^o

d = 20 m

.

Using formula : v² - u² = 2 a d

We can calculate the acceleration of the body :

Here, v² - u² = 2 a d

∴ (0)² - (10 m/s)² = 2 a (20 m)

∴ a = - 2.5 m/s²

^.

From the given free body diagram :

N = mg cosθ

And, F_net = mg sinθ + f_k = m a

∴ mg sinθ + μ_k N = m a

∴ mg sinθ + μ_k mg cosθ = m a

∴ g sinθ + μ_k g cosθ = a

∴ sinθ + μ_k cosθ = a / g

∴ sin(30) + μ_k cos(30) = (2.5) / (9.8)

∴ μ_k = 0.28