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 2.50 rev/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?

Answer 1

Answer:

$P = 57.6 N * 3.14 m/s = 180.956 W$

Explanation:

Data given

$F = 180 N$ represent the applied force by the wheel

$r = 0.2m$ represent the radius for the wheel

$w= 2.5 rev/s$ represent the rotational speed given

$\mu_k = 0.32$ represent the kinetic friction coeffcient given.

Solution to the problem

For this case we need to find first the spped with the usual units in m/s like this:

$v = \frac{2\pi R}{T} = 2\pi R w$

And we can replace the values given and we got:

$v = 2\pi (0.2 m) (2.5 rev/s) = 3.142 m/s$

Now we can calculate the friction force, for this case we assume that the normal force is the force exerted by the wheel and from the definition:

$F_k = \mu_k F = 0.32*180 N= 57.6 N$

Now we can calculate the power assuming that all the energy is converted to kinetic energy from the motor with the following formula:

$P = F v$

And we can replace with the values that we found:

$P = 57.6 N * 3.14 m/s = 180.956 W$

And that would be the energytransferred 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.