Physics, asked by deepikamsharma18, 1 month ago

A truck of mass 2000 kg changes its velocity from 18km/hr to 36km/hr in 10sec calculate work done by the truck and power exerted.​

Answers

Answered by Anonymous
2

Answer:

Work done by the truck is 75 × 10³ Joule and power exerted is 7.5 watt.

Explanation:

Given:

mass, m = 2000kg

Initial velocity, u = 18 km/hr

Final velocity, v = 36 km/hr

time, t = 10 sec

To Find:

(i) Work done by the truck.

(ii) Power exerted by the truck.

Formulas Used:

Work done, w = ∆KE = \sf \dfrac{m(v^2 - u^2)}{2}

Power, P = \sf \dfrac{work \ done}{time}

Solution:

(i) Work done by the truck = ?

Work done (w) = ∆KE = \sf \dfrac{m(v^2 - u^2)}{2}

w = \sf \dfrac{2000 \times (10^2 - 5^2)}{2}

\sf \dfrac{(100 - 25) \times 2000}{2}

\sf \dfrac{(100 - 25) \times \cancel{2000}}{\cancel{2}}

\implies \sf 75 \times 1000

\implies \sf 75 \times 10^3 \ J

{\pink{\bf{\therefore Work \ done \ by \ the \ truck \ is \ 75 \times 10^3 \ Joules.}}}

 \\

(ii) Power exerted by the truck = ?

Power (P) = \sf \dfrac{Work \ done}{time}

\implies \sf \dfrac{75}{10}

\implies \sf P = 7.5 watt

{\bf{\pink{\therefore Power \ exerted \ by \ the \ truck \ is \ 7.5 \ watt.}}}

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