raeganwalker5663 raeganwalker5663
  • 04-02-2021
  • Physics
contestada

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

Respuesta :

onyebuchinnaji
onyebuchinnaji onyebuchinnaji
  • 05-02-2021

Answer:

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;

[tex]W = K.E = \frac{1}{2} m(v^2-u^2)\\\\W = \frac{1}{2} \times 5(10^2-6^2)\\\\W = 160 \ J[/tex]

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

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