Question

A car of 1000kg increases its speed from 10m/s to 20m/s, While moving 500m up in a road inclined at an angle x to the horizontal where sin x is equal to 1/2.
There is a constant resistance to motion of 300N. Find the driving force exerted by the engine assuming that it is constant(g=10 meter per second squared)

Answer 1

We have:

$$v_f^2-v_i^2=2as$$

$$20^2-10^2=2(500)a\to a=0.3\frac{m}{s^2}$$

$$F=ma+mg\sin{\alpha}+F_r$$

$$F=1000(0.3+0.5(10))+300=5600\ N$$