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3 years ago

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

Physics

1 answer:

ahrayia [7]3 years ago

8 0

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;

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

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

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3 years ago

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Answer:

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c) -80.3 ft/s

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If y(5s) = - 160 ft:

-160 ft = -16.1 ft/s^2 * (5 s)^2 + Vo*(5s)

Solving for Vo

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To answer this question we must first know when the velocity became zero, at this time is when the ball was at its highest point.

v(t) = -32.2 ft/s^2 * t + Vo

t = Vo/32.2ft/s^2 = 1.5 s

And now, the highest point which the ball reached is given by:

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By means of the quadratic formula, we find that

t' = 4.00498 s ≈ 4 s

And the velocity at t = 4s is:

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