Question

5.3 two 30 kg children in a 20 kg cart are stationary at the top of a hill. They start rolling down the 80 m tall hill and they are travelling at 30 km h-1 when they reach the bottom. how muvh work was done on the cart by friction during it's travel down the hill

Answer 1

Answer:

60008.4 J

Explanation:

The mass of each kid = 30 kg

mass of the cart = 20 kg

The speed of the cart down the hill = 30 km/hr = 30 x 1000/3600 = 8.33 m/s

The height of the hill = 80 m

The potential energy of the boys at the top of the hill = mgh

where

m is the total mass of the kids and the cart = (30 x 2) + 20 = 80 kg

g is the acceleration due to gravity = 9.81 m/s^2

h is their height above the ground = 80 m (on the top of the hill)

substituting, we have

potential energy PE = 80 x 9.81 x 80 = 62784 J

At an instance at the bottom of the hill

their kinetic energy = $\frac{1}{2} mv^2$

where

v is their velocity = 8.33 m/s

m is their total mass = 80 kg

substituting, we have

kinetic energy KE = $\frac{1}{2}*80*8.33^2$ = 2775.6 J

Total work done on the cart is equal to the energy lost by the cart when it reached the bottom of the hill

work done by friction = PE - KE = 62784 - 2775.6 = 60008.4 J