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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

User Mkrakhin
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1 Answer

10 votes

Answer:

the net work done on the cart is 160 J.

Step-by-step 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;


W = K.E = (1)/(2) m(v^2-u^2)\\\\W = (1)/(2) * 5(10^2-6^2)\\\\W = 160 \ J

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

User Gngrwzrd
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