Question

A particle is moving with kinetic energy E, straight up an inclined plane with angle $\alpha$, the coefficient of friction being $\mu$. The work done against friction before the particle comes down to rest is

Answer 1

The work done against friction is $\frac{\mu E}{\tan \alpha + \mu}$.

Answer 2

When the particle moves up the inclined plane, the forces acting against its motion are:

1. Component of gravity along the incline: $mg \sin \alpha$
2. Kinetic friction force: $f_k = \mu N$. The normal force $N = mg \cos \alpha$. So, $f_k = \mu mg \cos \alpha$.

Let $d$ be the distance the particle travels up the incline before coming to rest.

The initial kinetic energy is $E$. The final kinetic energy is $0$ (comes to rest).

According to the work-energy theorem, the change in kinetic energy is equal to the net work done on the particle:

$\Delta E_k = W_{net}$

$0 - E = W_g + W_f$

Where $W_g$ is the work done by gravity and $W_f$ is the work done by friction.

Both forces oppose the motion, so their work done is negative:

$W_g = -(mg \sin \alpha) d$

$W_f = -(\mu mg \cos \alpha) d$

Substituting these into the work-energy equation:

$-E = -(mg \sin \alpha) d - (\mu mg \cos \alpha) d$

$-E = -mg d (\sin \alpha + \mu \cos \alpha)$

$E = mg d (\sin \alpha + \mu \cos \alpha)$

From the above equation, the distance $d$ is:

$d = \frac{E}{mg (\sin \alpha + \mu \cos \alpha)}$

The work done against friction is the magnitude of the work done by the friction force, which is $W_{against\_f} = f_k d$.

$W_{against\_f} = (\mu mg \cos \alpha) d$

Substitute the expression for $d$:

$W_{against\_f} = (\mu mg \cos \alpha) \left( \frac{E}{mg (\sin \alpha + \mu \cos \alpha)} \right)$

$W_{against\_f} = \frac{\mu E \cos \alpha}{\sin \alpha + \mu \cos \alpha}$

To simplify, divide the numerator and denominator by $\cos \alpha$:

$W_{against\_f} = \frac{\mu E}{\frac{\sin \alpha}{\cos \alpha} + \frac{\mu \cos \alpha}{\cos \alpha}}$

$W_{against\_f} = \frac{\mu E}{\tan \alpha + \mu}$