Question

A cart is set rolling across a level table, at the same speed on every trial. If it runs into a patch of sand, the cart exerts on the sand an average horizontal force of 6N and travels a distance of through the sand as it comes to a stop. If instead the cart runs into a patch of gravel on which the cart exerts an average horizontal force of 9N, how far into the gravel will the cart roll before stopping? (a) 9cm (b) 6cm (c) 4cm (d) 3cm (e) none of those answers.

Short answer

The correct option is (c)4cm.

Step-by-step solution

Given information

The average horizontal force exerted by the cart on the sand, ${\mathrm{f}}_{\mathrm{k}}=6\mathrm{N}$.

The distance of travel of the cart through the sand, $\mathrm{d}=6\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.06\mathrm{m}$.

The average horizontal force exerted on the cart by a patch of gravel ${{\mathrm{f}}^{\text{'}}}_{\mathrm{k}}=6\mathrm{N}$.

Work due to friction

When a body moves on a rough surface, a frictional force acts on the body, and the work is done on the body by the frictional force.

The mechanical energy loss of the body moving through a rough surface is converted into the work done in moving the body against friction.

The distance cart rolls before stopping 

In this system consisting of a rolling cart, the cart’s mechanical energy gets transformed into the work done against the friction from the surface. Balancing the loss of mechanical energy with the friction work,

$\begin{array}{c}\Delta E_{\mathrm{mech}}=f_{k}d\\ =6\mathrm{N}\times 0.06\mathrm{m}\times \frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\\ =0.36\mathrm{J}\end{array}$

Similarly, for the cart rolling through gravel,

${\mathrm{\Delta E}}_{\mathrm{mech}}=\mathrm{f}{\text{'}}_{\mathrm{k}}\mathrm{x}$

Here, x is the distance the cart rolls through gravel before stopping.

Put the values.

$\begin{array}{l}0.36\mathrm{J}=9\mathrm{N}\times x\times \frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\\ d=\frac{0.36\mathrm{J}}{9\mathrm{N}\times \frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}}\\ d=0.04\mathrm{m}\left(\frac{100cm}{1m}\right)\\ d=4\mathrm{cm}\end{array}$

Hence, the distance cart rolls through gravel before stopping is 4cm, and the correct option is (c) 4cm.