Question

a body starts moving up an inclined plane of angle 30 degree with an initial kinetic energy of 20J . the coefficient of friction is 1/root 3. the work done against friction before the body comes to rest is
1) 5J
2) 7J
3) 10J
4) 15J

Answer 1

answer : option (1) 5J

explanation : inclination angle of plane , Φ = 30°

coefficient of friction, μ = 1/√3

initial kinetic energy of body , K.E = 20J

let mass of body is m

so, K.E = 1/2 mu²

or, 20 = mv²

or, v = √{20/m}

using formula, v² = u² + 2as

or, 0 = (20/m) + 2as .....(1)

here, a is acceleration. let's find it.

body moves upward, so net force = -(weight + frictional force)

or, ma = -(mgsinΦ + μmgcosΦ)

or, a = -g(sinΦ + μ cosΦ) [ downward direction along plane ]

putting a in equation (1),

20/m =2g(sinΦ + μ cosΦ)s

or, s = 20/2g(sinΦ + μ cosΦ) .....(2)

now, workdone due to frictional force, W = (μmgcosΦ)s

from equation (2),

= 20μgcosΦ/2g(sinΦ + μ cosΦ)

= 10μcosΦ/(sinΦ + μ cosΦ)

= 10 × 1/√3 × cos30°/(sin30° + 1/√3 × cos30°)

= 5/(1/2 + 1/2)

= 5 J

hence, option (1) is correct choice