Solveeit Logo

Question

Question: Find the work done by the frictional force in drawing a circle of radius \(r\) by a pencil of neglig...

Find the work done by the frictional force in drawing a circle of radius rr by a pencil of negligible mass with a normal pressing force NN (coefficient of friction μ\mu ).

Explanation

Solution

In this question, the concept of friction force is applied, that is the friction force is the product of the coefficient of friction and the normal force exerted on the body or the object. First, draw the diagram using all the information given in the question and then find out the work done by the frictional force in drawing a circle of radius and discuss the derivation.

Complete step by step solution:
As per the given question first, we will draw the diagram as per the given situation,

The motion of the pencil is in the zz direction. The given radius of the circle is rr. The friction of motion will be tangentially. So, we have to derive the work done at dldl length.
We know that the work done at length dldl can be written as,
dw=fk.dldw = - {f_k}.dl
Here, the friction force is fk{f_k}.
As we know that the frictional force is always opposing the relative motion of the body so it is negative.
Now, we integrate the above equation as,
dw=w\int {dw} = w
w=fk.dl\Rightarrow w = - \int {{f_k}.dl}
As we know that the friction force is the product of the coefficient of friction and the normal force exerted on the body or the object so,
fk=μN{f_k} = \mu N
Where, μ\mu is coefficient of friction and NN is the pressing force.
Now we will substitute the value of friction force in the work done equation as,
w=μNdlw = - \int {\mu Ndl}
The total length of the circle is 2πr2\pi r. So to find out the total work done at length dldl we have to integrate from 00 to 2πr2\pi r, so the above equation become,
w=02πrμNdlw = - \int\limits_0^{2\pi r} {\mu Ndl}
We integrate the above equation as,
w=μN[2πr0]w = - \mu N\left[ {2\pi r - 0} \right]
w=2πμrN\Rightarrow w = - 2\pi \mu rN
Thus, the total work done by the frictional force is 2πrN - 2\pi rN.

Note: When a body is dragged along a rough surface, the frictional force will be acting in the direction opposite to the displacement of the body. And the angle between the displacement of the body and frictional force will be 180{180^ \circ }. Hence the total work done by the frictional force will be negative.