A 25.0kg pickle is accelerated from rest through a distance of 6.0m in 4.0s across a level floor. If the friction force between the pickle and the floor is 3.8N, how much work is required to move the object?

Answer 1

d = Vo*t + 0.5a*t^2 = 6 m.
0 + 0.5a*4^2 = 6
8a = 6
a = 0.75 m/s^2.

Fp = m*g = 25kg * 9.8N/kg = 245 N. = Force of pickle.

Fap-Fp-Fk = m*a
Fap-245-3.8 = 25*0.75
Fap = 18.75+245+3.8 = 267.6 N. = Force
applied.

Work = Fap * d = 267.6 * 6 = 1605 Joules

Answer 2

d = Vo*t + 0.5a*t^2 = 6 m.
0 + 0.5a*4^2 = 6
8a = 6
a = 0.75 m/s^2.

Fnet=m*a
Fnet=25*0.75=18.75 N

Fnet=Fap-Ff
Fap=Fnet+Ff
Fap=18.75+3.8=22.55 N

W=Fap*d
W=22.55*6=135.3 J