A girl pulls a box of mass m = 5 kg across a floor with a constant horizontal fo

Question

A girl pulls a box of mass m = 5 kg across a floor with a constant horizontal force F = 23 N. Initially the block is at rest. For the first d_1 = 3 m, there is no friction between the box and the floor. For the next d_2 = 6 m the coefficient of friction between the box and the floor is mu = 0.1. What is the work done on the box by the girl in moving the box over the distance d_1 + d_2? W_g = What is work done on the box by friction in moving the box over the distance d_1 + d_2?W_floor = What is the final speed of the box (after being pushed to d_1 = d_2)? V_f =

Explanation / Answer

Here,

m = 5 kg

F = 23 N

d1 = 3 m

d2 = 6 m

u = 0.1

1)

work done by the girl = F * (d1 + d2)

work done by the girl = 23 * (3 + 6)

work done by the girl = 207 J

2)

Wfloor = -u * m* g * d2

Wfloor = - 0.1 * 5 * 9.8 * 6

Wfloor = - 29.4 J

the work done by the friction in moving the box over the distance is -29.4 J

3)

let the final speed is v

Using Work energy theorum

0.5 * 5 * (v^2 - u^2) = net work done

0.5 *5 * (v^2) = 207 - 29.4

v = 8.43 m/s

the final speed of the box is 8.43 m/s

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