Question

A particle is thrown with kinetic energy k straight
up a rough inclined plane of inclination a and
coefficient of friction u. The work done against
friction before the particle comes to rest is
kusina
(1) cosa + u sina
kucosa
cosa + u sina
kusina
sina +ucosa
kucosa
sina +ucosa​

Answer 1

it is given that,

angle of inclination of plane = α

coefficient of friction = μ

Kinetic energy = K

from law of conservation of energy.

kinetic energy of particle = workdone by frictional force + potential energy when particle comes to rest.

so, first of all, we have to find the height, reached by particle.

initial velocity of particle, u = √{2K/m}, where m is mass of particle.

ma = mgsinα + μmgcosα

a = (gsinα + μgcosα) [ opposite to motion of particle along plane ]

now use formula, v² = u² + 2as

0 = 2K/m + 2{-(gsinα + μgcosα)}s

or, s = K/{m(gsinα + μgcosα)}

then, potential energy = mgh

where, h = ssinα

= Ksinα/(sinα + μcosα)

now, work done by the frictional force = kinetic energy - potential energy

= K - Ksinα/(sinα + μcosα)

= K[(sinα + μcosα - sinα)/(sinα + μcosα) ]

= Kμcosα/(sinα + μcosα)