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Question: The work done by the frictional force on a pencil in drawing a complete circle of radius \(r = \dfr...

The work done by the frictional force on a pencil in drawing a complete circle of

radius r=1πr = \dfrac{1}{\pi } metre on the surface by a pencil of negligible mass with a normal

pressing force N=5N(μ=0.5)N = 5N(\mu = 0.5)

(A) +4J + 4J

(B) 3J - 3J

(C) 2J - 2J

(D) 5J - 5J

Explanation

Solution

In order to solve the problem we need to know about what is work done by frictional force and knowledge about normal force , direction of normal force , direction of frictional force.

Work done by frictional force is

dwdw

$$$$dw=Fk.dldw = \overrightarrow {{F_k}} .dl

Here Fk\overrightarrow {{F_k}} is the frictional force opposite to the applied force

Due to opposite in direction we can say that

dw=Fk.dldw = - \overrightarrow {{F_k}} .dl

Given:

a pencil in drawing a complete circle of radius r=1πr = \dfrac{1}{\pi } metre on the surface by a pencil of negligible mass with a normal pressing force N=5NN = 5N

and μk=0.5{\mu _k} = 0.5(coefficient of kinetic friction)

To find the work done by the frictional force on a pencil

Complete step by step solution:

As per the given question first, we will draw the diagram as per the given situation,

Let the frictional force applied be Fk\overrightarrow {{F_k}} ​​

Fk\overrightarrow {{F_k}} acts tangentially opposite to the motion of the pencil.

For elementary displacement dldl

The work done by Fk\overrightarrow {{F_k}} is dwdw

dw=Fk.dl=FK.dldw = \overrightarrow {{F_k}} .dl = - \overrightarrow {{F_K}} .dl

Integrating on both side, we get

0Wdw=02πrFK.dl\int_0^W {dw} = - \int_0^{2\pi r} {\overrightarrow {{F_K}} .dl}

0Wdw=μkN02πrdl\int_0^W {dw} = - {\mu _k}N\int_0^{2\pi r} {dl} as we know (FK=μkN)(\overrightarrow {{F_K}} = {\mu _k}N)

W=μkN2πrW = - {\mu _k}N2\pi r

Put value of μk,r,N{\mu _k},r,N

W=0.5×5×2×π×1πW = -0.5 \times 5 \times 2 \times \pi \times \dfrac{1}{\pi }

W=5JW = - 5J is the work done by the frictional force on a pencil in drawing a complete circle.

So the correct answer is (d).

Note: We see friction in our day to day activities. Following are the ten examples of friction in daily life:

1. Walking

2. Writing

3. Scating

4. Lighting a matchstick

5. Driving of the vehicle on a surface

6. Applications of breaks in the vehicle to stop it

7. Flight of aeroplanes

8. Drilling a nail into the wall

9. The dusting of the carpet by beating it with a stick

10. Sliding on a garden slide