Question

problem 2: you push your desk horizontally in a circular arc of radius r = 3 m, as shown on the figure. the mass of the desk is m = 15 kg, and the coefficient of kinetic friction between the floor and the desk is μk = 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 normal force

The normal force $N$ on a horizontal - surface is equal to the weight of the object. Since $F_g = mg$ and $N = F_g$, where $m = 15$ kg and $g=9.8$ m/s², we have $N=mg = 15\times9.8=147$ N.

Step2: Calculate the frictional force

The frictional force is given by $F_f=\mu_kN$. Substituting $\mu_k = 0.3$ and $N = 147$ N, we get $F_f=\mu_kN=0.3\times147 = 44.1$ N.

Step3: Calculate the distance along the circular arc

The length of a semi - circular arc is $s=\pi R$, with $R = 3$ m, so $s=\pi\times3=3\pi$ m.

Step4: Calculate the work done by friction along the circular arc

The work - done formula is $W = F\cdot d\cdot\cos\theta$. The frictional force is always opposite to the direction of motion, so $\theta = 180^{\circ}$ and $\cos\theta=- 1$. Then $W_{arc}=-F_fs=-44.1\times3\pi\approx - 415.5$ J.

Step5: Calculate the work done by friction along a straight line

The straight - line distance between $A$ and $B$ for a semi - circle of radius $R$ is $d = 2R=6$ m. Using $W = F\cdot d\cdot\cos\theta$ with $\theta = 180^{\circ}$ and $F_f = 44.1$ N, we get $W_{line}=-F_f\times2R=-44.1\times6=-264.6$ J. So the answer changes.

Answer:

The work done by friction along the circular arc is approximately $-415.5$ J. The answer would change if the path was a straight line.