Question

Zamboni has broken down and is being pulled across the ice(consider a frictionless) The workers pull the Zamboni with a rope which makes an angle of 10.0^• with The horizontal force. The force of tension in the rope due to the pole of the two workers is 47.9 and if the repair bay is 40.0 M away will the workers be able to make it if they expended 2.00 x10³ J of work?

Answer 1

Mass of zamboni

$2Tcos\theta=mg$

$M=\frac{2Tcos\theta}{g}=\frac{2\times 47.9cos10}{9.8}$

$=9.627kg$

Required time

$W=2\times 10^3J$

$W=Fd$

$F=\frac{w}{d}=\frac{2\times10^3}{40}=50N$

$50=9.627\times\frac{40}{t^2}=t=\frac{\sqrt{9.627\times40}}{50}=2.875$

Time taken

The net force forward will be the sum of forward components of the two ropes

Each rope , $Fx=Fcos\theta=47.9cos10$

$=47.17N$

So net forward force, $F_{forward}=2\times 47.17N=94.34N$

$F=\frac{m\triangle V}{\triangle t}$

$94.34=m\times\frac{40.0}{t^2}=94.34=9.627\times\frac{40.0}{t}$

$t=\sqrt{\frac{9.627\times 40}{94.34}}=\sqrt{4.08seconds}=2.02 sec$

$\therefore$ The workers will make it to the repair bay