Question

what is a reciprocal of 123÷18/20​

Answer 1

Answer:

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+umgcosak=mgsina+umgcosa ...(1)

w_{friction}=umgcosaw

friction

=umgcosa ...(2)

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

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

k

w

friction

=

mgsina+umgcosa

umgcosa

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

friction

=

sina+ucosa

k×ucosa

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

friction

=

sina+ucosa

ukcosa

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

\frac{ukcosa}{sina+ucosa}

sina+ucosa

ukcosa