At time t_(1) a cart with moss m being pushed on a straight, level track has speed o. A short time later at l_(2) the cart has speed So Which of the following expressions is equal to the average power delivered to the cart between time t_(1) and t_(2) A 4mv^2 D (9)/(2)mv^2 C (4mv^2)/(t_(2)-t_(1)) D (4mv^2)/(2(a-t_(1)))

Question

At time t_(1) a cart with moss m being pushed on a straight, level track has speed o. A short time later at l_(2) the cart has speed So Which of the following expressions is equal to the average power delivered to the cart between time t_(1) and t_(2) A 4mv^2 D (9)/(2)mv^2 C (4mv^2)/(t_(2)-t_(1)) D (4mv^2)/(2(a-t_(1)))

Answer 1

The average power delivered to the cart is the change in kinetic energy divided by the change in time.The kinetic energy at time t₁ is (1/2)mv²The kinetic energy at time t₂ is (1/2)m(3v)² = (1/2)m(9v²) = (9/2)mv²The change in kinetic energy is:(9/2)mv² - (1/2)mv² = (8/2)mv² = 4mv²The change in time is t₂ - t₁Therefore, the average power is:(Change in Kinetic Energy) / (Change in time) = 4mv² / (t₂ - t₁)So the correct answer is C.

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