Question

A cart that weighs 70 N is pushed 50 meters from rest with 1400 N. What is the final
velocity of the
cart?

Answer 1

Answer:

$v_f \approx 4.43\cdot10^2 m/s$

Explanation:

We can assume that all the work happening during the 50m push contributes to the cart kinetic energy, due to the kinetic energy theorem. It means that, supposing the force is applied in the direction of motion, the cart gains an energy of $W= \vec F \cdot \vec x = 1400 \cdot 50 \cdot 1 = 70 kJ$. This energy allows you to determine the kinetic energy of the cart, thus its velocity.

The only missing data is the mass of the cart, which is easily obtainable from it's weight, by dividing it by the gravitational acceleration $g= 9.81 m/s^2$, obtaining a mass of $70/ 9.81 = 7.13 kg$

Now we have everything, and from the KE theorem

$W = \Delta K = \frac12m v_f^2 -\frac12mv_i^2$ where, replacing and solving for whatever we have left:

$7\cdot 10^5 = \frac12 (7.13)\cdot v_f^2 -\frac12 (7.13)\cdot0^2\\ 7\cdot 10^5 = \frac12 (7.13)\cdot v_f^2\\v_f^2 = 1.96\cdot 10^5 \rightarrow v_f \approx 4.43\cdot10^2 m/s$

Feel free to double check calculations, as usual