Question

problem 2:
you push your desk horizontally in a circular arc of radius ( r = 3 mathrm{~m} ), as shown
on the figure. the mass of the desk is ( m = 15 mathrm{~kg} ), and the coefficient of kinetic
friction between the floor and the desk is ( mu_{k}=0.3 ). find the work done by
friction as you go from ( a ) to ( b ). would the answer change if the path you took
was a straight line?

Explanation:

Step1: Calculate the force of kinetic friction

The force of kinetic friction is given by \(F_f=\mu_k N\). Since the desk is on a horizontal surface, \(N = mg\) (where \(g = 9.8\ m/s^2\)). So \(F_f=\mu_kmg\).
Substituting the values \(m = 15\ kg\), \(\mu_k=0.3\), and \(g = 9.8\ m/s^2\), we get \(F_f=0.3\times15\times9.8 = 44.1\ N\).

Step2: Calculate the distance along the circular arc

The length of a circular arc is \(s=\frac{\pi R}{2}\) (since it's a semic - arc, and for a full circle \(s = 2\pi R\), here the angle \(\theta=\pi\) radians). Given \(R = 3\ m\), \(s=\frac{\pi\times3}{2}\approx4.71\ m\).

Step3: Calculate the work done by friction

The work done by a force \(W=\vec{F}\cdot\vec{s}=-F_fs\) (negative because the force of friction is opposite to the direction of motion).
Substituting \(F_f = 44.1\ N\) and \(s\approx4.71\ m\), \(W=-44.1\times4.71=-207.711\ J\).

Step4: Analyze the case of a straight - line path

If the path is a straight line, the distance \(s'=\sqrt{2}R\approx\sqrt{2}\times3\approx4.24\ m\).
The work done \(W'=-F_fs'=- 44.1\times4.24=-187\ J\) (approx). So the answer changes because the distance \(s\) (over which the non - conservative force of friction acts) changes.

Answer:

The work done by friction along the circular arc is approximately \(-208\ J\). Yes, the answer would change if the path was a straight line.