# physics helppp?

If a 65 kg person accelerates uniformly from a standing start to a sprint in 2.7 s over 10.0 m, what was their work done while accelerating?

### 3 Answers

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- oubaasLv 74 months ago
acceleration a = 2d/t^2 = 2*10/2.7^2 = 2.74 m/sec^2

V^2 = 2*a*d = 2.74*2*10 = 54.9 m^2/sec^2

Work = KE = 65/2 * 54.9 = 1.8*10^3 joule

- FiremanLv 74 months ago
By s = ut + 1/2at^2

=>10 = 0 + 1/2 x a x (2.7)^2

=>a = 2.74 m/s^2

By W = F x s

=>W = (ma) x s

=>W = 65 x 2.74 x 10

=>W = 1783.26 J

- fmcarajoLv 74 months ago
v² = vo² + 2ad

v² = 0 + 2a(10)

v² = 20a (1)

v = vo + at

v = 0 + at

v = 2,7a

v² = 2,7²a² (2)

(1) in (2)

20a = 2,7²a²

7,29a² - 20a = 0

a1 = 2,74 m/s² (Correct)

a2 = 0 (It doesn't exist)

v = 2,7(2,74) = 7,40 m/s

Ek = 0,5mv²

Ek = 0,5(65)(7,40²) = 1778,74 j

Review it!

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