Answer:
(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.
