Question

A cart of mass 20 kg at rest is to be dragged at a speed of 18 km/hr. If the co-efficient of friction between the cart and the ground is 0.1, what is the minimum force required to drag a cart to a distance of 10 m? (g = 10 m/s^2)

Please answer fast.

Answer 1

18 km/hr = 5m/s

using eq v^2 - u^2 = 2as we get

a = 1.25 m/s^2

now f = µmg + ma

= 32.5 N

Answer 2

Given, v = 18 km/h = 18×1000 3600=5 m/s2as=v2−u2⇒a=v2−u22s=2520=1.25 m/s2
where a is the acceleration of the cart

The minimum force required to drag the cart to distance 10 m = μmg+ma
F=μmg+ma⇒F=0.1×20×10+10×1.25 =32.5 N