A 5 .0kg cart is moving horizontaly at 6 .0m/s in order to change its speed to 10m/s the net work done on the cart must be

the net work done on the cart is 160 J.

Explanation:

Given;

mass of the cart, m = 5.0 kg

initial velocity of the cart, u = 6 m/s

final velocity of the cart, v = 10 m/s

The net work done on the cart is equal to change in average kinetic energy of the cart;

Therefore, the net work done on the cart is 160 J.

your answer would be fusion

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6