Question

A grinding machine whose wheel has a radius of 1/11 is rotating with angular speed at $2.5rev/\sec$. A tool to be sharpened is held against the wheel with a force of $40N.$If the coefficient of friction between the tool and the wheel is 0.2, power required is
(A) $40W$
(B) $4W$
(C) $8W$
(D) $10W$

Answer 1

calculate the friction force using the normal and coefficient of friction once you know the value of friction then find the velocity of the point of contact and then take dot product of the friction force and velocity find the power.

Complete step by step answer:
The tool is held against the wheel with a force of $40N$ it means normal force acting on the wheel will be $40N$ as both forces must be balanced
Now coefficient of friction is given so we can calculate friction
We know that,$f = \mu N$ where f is friction force,$\mu \$ is coefficient of friction and N is normal force
Therefore, putting values we have,
$f = 0.2 \times 40 = 8N$
Now we will calculate the velocity of a point on the rim of the wheel
We have angular velocity $\omega = 2.5rev/s$converting it into rad/s we have
$\omega = 2.5 \times 2 \times \pi rad/s = 5\pi rad/s$
We know that $v = \omega r$where r is the radius
So $v = 5\pi \times \dfrac{1}{{11}} = \dfrac{{5\pi }}{{11}}$
Now direction of friction and velocity will be opposite as friction opposes the motion. So,
Power $P = f.v \Rightarrow f \times v \times \cos {180^ \circ } = 8 \times \dfrac{{5\pi }}{{11}} \approx 10W$
Therefore power required is $10W$.

Hence, Option-D is correct.

Note: In such type of questions you need to visualize the force carefully here for example force required to held against wheel is normal force because tangential force be used in sharpening the object which we intended to do and once you know the force then you can take the dot product with velocity to find the power asked.