A force of 6.7 N acts on a 30 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.

(a) 0.748 J

(b) 2.245 J

(c) 3.74 J

(d) 4.482 W

Explanation:

(a) Work done  W = Force × distance

W = F×d,

Where d = 1/2(at²)

Therefore,

W =1/2(F×at²) Equation 1

Where a = acceleration, t = time.

But,

a = F/m Equation 2

Where m = mass.

Substitute equation 1 into equation 2

W = 1/2(F²t²/m) Equation 3

Given: F = 6.7 N, t = 1 s, m = 30 kg

Substitute into equation 3

W₁ = 1/2(6.7²×1²/30)

W = 0.748 J.

(b) Similarly,

The work done in the second seconds is

Where t₂ = 2 s

W₂ = 1/2(F²t₂²/m)- W₁

W = 1/2(6.7²×2²/30)-0.748

W = 2.245 J

(c) The work done in the third seconds is

Where t₃ = 3 s

W₃ = 1/2(F²t₃²/m)-(W₂+W₃)

W = 1/2(6.7²×3²/30)-(2.993)

W = 3.74 J.

(d) P = Fv Equation 4

Where v = velocity.

and,

v = at Equation 5

Substitute equation 5 into equation 4

P = Fat Equation 6

Given: F = 6.7 N, a = 6.7/30 = 0.223 m/s², t = 3 s

Substitute into equation 6

P = 6.7×0.223×3

P = 4.482 W.

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