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A body of mass 0.1 kg falling freely under gravity takes 10 s to reach the ground. Calculate the kinetic energy and potential energy of the body when it has travelled for first 6 s. Given, g = 9.8 m/ s²​

Sagot :

Answer: KE = 172.872 J

PE = 307.328 J

Explanation:

For KE: acceleration = change in velocity / time → 9.8 = final velocity - 0 (because initial velocity is zero) / 6s → final velocity after 6s = 58.8 m/s.

Now, plug this into: KE = 1/2(m)(v^2) → KE = 1/2(.1)(58.8^2) = 172.872 J.

For PE: Find the velocity after 10 s: acceleration = change in velocity / time → 9.8 = final velocity - 0 (because initial velocity is zero) / 10s → final velocity after 10s = 98 m/s.

Now, find the total height: (v final)^2 = (v initial)^2 + 2*a*d → 98^2 = 0^2 + 2*9.8*d → d = total height = 490 m.

Now, find the change in height after 6 s: (v final)^2 = (v initial)^2 + 2*a*d → 58.8^2 = 0^2 + 2*9.8*d → d = change in height after 6 s → 176.4 m

Now, subtract the change in height from the total height to get its height after 6s: 490-176.4=313.6m

PE = m*g*h → Pe=.1*9.8*313.6 = 307.328 J :)

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