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
ku sina
(1) cosa + u sina
kucosa
cosa + u sina
(3) sinat
kusina
sina +ucosa
ku cosa
sina + u cosa​

Answer 1

Answer:

Explanation:

Pic attached

Hope you understand!!

Answer 2

Given:

Kinetic energy=k

Inclination angle=a

Coefficient of friction=u

To find:

Work done against friction before the particle comes to rest

Solution:

Kinetic energy=Work done due to gravity+ work done  due to friction

$k=mgsina+umgcosa$...(1)

$w_{friction}=umgcosa$...(2)

Equation (2) divided by (1) then, we get

$\frac{w_{friction}}{k}=\frac{umgcosa}{mgsina+umgcosa}$

$w_{friction}=\frac{k\times ucosa}{sina+ucosa}$

$w_{friction}=\frac{ukcosa}{sina+ucosa}$

Hence, the work done against friction before the particle comes to rest is

$\frac{ukcosa}{sina+ucosa}$