Question

A 98.0 N grocery cart is pushed 12.0 m along an aisle by a shopper who exerts a constant horizontal force of 40.0 N. If all frictional forces are neglected and the cart starts from rest, what is the grocery cart's final speed?

Answer 1

Answer:

9.8 m/s

Explanation:

The work done by the force pushing the cart is equal to the kinetic energy gained by the cart:

$W=K_f -K_i$

where

W is the work done

$K_f$ is the final kinetic energy of the cart

$K_i$ is the initial kinetic energy of the cart, which is zero because the cart starts from rest, so we can write:

$W=K_f$

But the work is equal to the product between the pushing force F and the displacement, so

$W=Fd=(40.0 N)(12.0 m)=480 J$

So, the final kinetic energy of the cart is 480 J. The formula for the kinetic energy is

$K_f=\frac{1}{2}mv^2$ (1)

where m is the mass of the cart and v its final speed.

We can find the mass because we know the weight of the cart, 98.0 N:

$m=\frac{F_g}{g}=\frac{98.0 N}{9.8 m/s^2}=10 kg$

Therefore, we can now re-arrange eq.(1) to find the final speed of the cart:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(480 J)}{10 kg}}=9.8 m/s$