A girl pulls a box of mass m = 2 kg across a floor with a constant horizontal fo

Question

A girl pulls a box of mass m = 2 kg across a floor with a constant horizontal force F = 23 N. Initially the block is at rest. For the first d1 = 5 m, there is no friction between the box and the floor. For the next d2 = 3 m the coefficient of friction between the box and the floor is µ = 0.2.

1) What is the work done on the box by the girl in moving the box over the distance d1 + d2?

Wg =

2) What is work done on the box by friction in moving the box over the distance d1 + d2?

Wfloor =

3) What is the final speed of the box (after being pushed to d1 + d2)?

vf =

Explanation / Answer

Normal force =N=mg= 2*9.8

Friction uN = 0.2*2*9.8 =3.92N

Work done W=F*d

During d1

W1= 23*5= 115J

Fnet during d2= 23-3.92=19.08

W2=19.08*3=57.24J

1) Wg= W1 + W2= 115+57.24=172.24J

2) Wf =f*d2=3.92*3=11.76J

3) Using work energy theorem

W=KE

172.24=0.5*2*vf2

vf= 13.124m/s

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