# A cart of mass 1000 kg is pulled by a horse of 200kg. The coefficient of friction between them and ground is 0.2. Calculate the force required to produce acceleration of 2 m/s^2 in the cart??'

Jul 16, 2018

${F}_{\text{pull}} = 3960 N$

#### Explanation:

The net force required to give the cart acceleration of $2 \frac{m}{s} ^ 2$ is given by

${F}_{\text{net}} = m \cdot a = 1000 k g \cdot 2 \frac{m}{s} ^ 2 = 2000 k g \cdot \frac{m}{s} ^ 2 = 2000 N$

The friction between the cart and the ground will be in the opposite direction of the pull on the cart.

The friction, ${F}_{f}$, will be

${F}_{f} = {\mu}_{k} \cdot N = 0.2 \cdot m \cdot g = 0.2 \cdot 1000 k g \cdot 9.8 \frac{m}{s} ^ 2$

${F}_{f} = 1960 k g \cdot \frac{m}{s} ^ 2 = 1960 N$

Therefore, the $2000 N$ net force is the resultant of the 2 forces on the cart.

#F_"net" = 2000 N = F_"pull" - F_f = F_"pull" - 1960N

$2000 N = {F}_{\text{pull}} - 1960 N$

${F}_{\text{pull}} = 2000 N + 1960 N = 3960 N$

... Assuming level ground.

I hope this helps,
Steve