A force of 62 N acts on a 35kg object for 10.0s. What is the object change in velocity?

A force of 62 N acting on a 35 kg object applies an acceleration a of

62 N = (35 kg) a

a = (62 N) / (35 kg) ≈ 1.8 m/s²

If v₀ is the object's initial velocity, then after 10.0 s it attains a final velocity v of

v = v₀ + a t

v = v₀ + (1.8 m/s²) (10.0 s)

v = v₀ + 18 m/s

so that the change in velocity is

∆v = v - v₀ = 18 m/s

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

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