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Question: A body is moving up an inclined plane of angle θ with an initial kinetic energy k. The coefficient...

A body is moving up an inclined plane of angle θ with an initial kinetic energy k. The coefficient of friction between the plane and body is μ . The work done against friction before the body comes to rest is

A

μk

B

k

C

k / (tanθ - μ)

D

μk / (tanθ + μ)

Answer

μk / (tanθ + μ)

Explanation

Solution

The problem asks for the work done against friction when a body moves up an inclined plane and comes to rest. We can use the work-energy theorem to solve this.

1. Identify Forces and Work Done:

When the body moves up the inclined plane, the forces opposing its motion are:

  • Component of gravity along the incline: Fg=mgsin⁡θF_g = mg \sin \theta
  • Kinetic friction force: fk=μNf_k = \mu N. The normal force N=mgcos⁡θN = mg \cos \theta. So, fk=μmgcos⁡θf_k = \mu mg \cos \theta.

Let dd be the distance the body travels up the incline before coming to rest.

The work done by gravity (WgW_g) and friction (WfW_f) are negative as they oppose the motion:

  • Wg=−(mgsin⁡θ)dW_g = - (mg \sin \theta) d
  • Wf=−(μmgcos⁡θ)dW_f = - (\mu mg \cos \theta) d

2. Apply Work-Energy Theorem:

The initial kinetic energy is Ki=kK_i = k. The final kinetic energy is Kf=0K_f = 0 (body comes to rest).

According to the work-energy theorem, the change in kinetic energy equals the net work done:

ΔK=Wnet\Delta K = W_{net}

Kf−Ki=Wg+WfK_f - K_i = W_g + W_f

0−k=−(mgsin⁡θ)d−(μmgcos⁡θ)d0 - k = -(mg \sin \theta) d - (\mu mg \cos \theta) d

−k=−mgd(sin⁡θ+μcos⁡θ)-k = -mg d (\sin \theta + \mu \cos \theta)

k=mgd(sin⁡θ+μcos⁡θ)k = mg d (\sin \theta + \mu \cos \theta)

3. Determine the Distance Traveled (d):

From the work-energy equation, we can find the distance dd:

d=kmg(sin⁡θ+μcos⁡θ)d = \frac{k}{mg (\sin \theta + \mu \cos \theta)}

4. Calculate Work Done Against Friction:

The work done against friction is the magnitude of the work done by the friction force, which is Wagainst_friction=fkdW_{against\_friction} = f_k d.

Wagainst_friction=(μmgcos⁡θ)dW_{against\_friction} = (\mu mg \cos \theta) d

Substitute the expression for dd:

Wagainst_friction=(μmgcos⁡θ)(kmg(sin⁡θ+μcos⁡θ))W_{against\_friction} = (\mu mg \cos \theta) \left( \frac{k}{mg (\sin \theta + \mu \cos \theta)} \right)

Wagainst_friction=μkcos⁡θsin⁡θ+μcos⁡θW_{against\_friction} = \frac{\mu k \cos \theta}{\sin \theta + \mu \cos \theta}

To simplify, divide the numerator and denominator by cos⁡θ\cos \theta:

Wagainst_friction=μkcos⁡θcos⁡θsin⁡θcos⁡θ+μcos⁡θcos⁡θW_{against\_friction} = \frac{\frac{\mu k \cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta} + \frac{\mu \cos \theta}{\cos \theta}}

Wagainst_friction=μktan⁡θ+μW_{against\_friction} = \frac{\mu k}{\tan \theta + \mu}

The work done against friction before the body comes to rest is μktan⁡θ+μ\frac{\mu k}{\tan \theta + \mu}.