Question

A metal tool is sharpened by being held against the rim of a wheel on a grinding machine by a force of 180 N. The frictional forces between the rim and the tool grind off small pieces of the tool. The wheel has a radius of 20.0 cm and rotates at 2.50rev/s . The coefficient of kinetic friction between the wheel and the tool is 0.320. At what rate is energy being transferred from the motor driving the wheel to the thermal energy of the wheel and tool and to the kinetic energy of the material thrown from the tool?

Short answer

The rate at which energy is being transferred from the motor driving the wheel to the thermal energy of the wheel and tool, and to the kinetic energy of the material thrown from the tool is equal to 181 W.

Step-by-step solution

Given data:

The force acting on the metal tool to hold it on the wheel, N = 180 N

Radius of the wheel, R = 20.0 cm = 0.20 m

Frequency of the wheel, f = 2.50 rev /s

The coefficient of kinetic friction between the wheel and the tool, ${\mu }_k=0.320$

To understand the concept:

From the frequency of the wheel, find the velocity of the wheel. Using normal force acting on the tool to hold it on the wheel and coefficient of friction, find the frictional force.

Then as power dissipated by the frictional force is equal to the power supplied by the motor, find the rate at which energy is being transferred from the motor driving the wheel to the thermal energy of the wheel and tool and to the kinetic energy of the material thrown from the tool using the relation between power, force, and velocity.

Formulae:

The frictional force is,

$F_k={\mu }_{k}N$

Here, ${\mu }_k$is the coefficient of friction and N is the normal force.

The power is define by,

P = Fv

Here, F is the force and v is the velocity.

Calculate the rate at which energy being transferred from the motor driving the wheel to the thermal energy of the wheel:

Velocity of the wheel is,

$$\begin{gathered} v=\frac{\mathit{2}\pi R}{T} \\ \mathit{}\mathit{}\mathit{}\mathit{=}\mathit{2}\pi Rf \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} v=2\times 3.14(0.20\mathrm{m})(2.50\mathrm{rev}/\mathrm{s}) \\ =3.14\mathrm{m}/\mathrm{s} \end{gathered}$$

The frictional force acting on tool is,

$$\begin{gathered} f_k={\mu }_{k}N \\ =0.320(180\mathrm{N}) \\ =57.6\mathrm{N} \end{gathered}$$

The rate at which energy is being transferred from the motor driving the wheel to the thermal energy of the wheel and tool and to the kinetic energy of the material thrown from the tool is nothing but the power supplied by the motor. This is equal to the power dissipated by the frictional force.

Hence,

$$\begin{gathered} P=f_{k}v \\ =(57.6\mathrm{N})(3.14\mathrm{m}/\mathrm{s}) \\ =180.86\mathrm{W} \\ =181\mathrm{W} \end{gathered}$$

Calculate the kinetic energy of the material thrown from the tool:

Therefore, the rate at which energy is being transferred from the motor driving the wheel to the thermal energy of the wheel and tool and to the kinetic energy of the material thrown from the tool is equal to 181 W .