Achild pushes horizontally on a box of massmwhich moves with constant speedvacrossa horizontal floor. the coefficient of frictionbetween the box and the floor isμ.

at what rate does the child do work on thebox?

1.p=μ m g

2.p=m g v

Complete Question:

A child pushes horizontally on a box of mass m which moves with constant speed across a horizontal floor. The coefficient of friction between the box and floor is u. At what rate does the child do work on the box? g= gravity

A:  umgv, mgv, v/umg, umg/v, umv^2 2.

1. μ*m*g*v

Explanation:

As the box is moving at constant speed, in the horizontal plane, we have two forces present, that must  be equal and opposite each other so the net horizontal force be zero: the force that the child does and the friction force.

⇒ Fc = Ff = μ*N (by definition of kinetic friction) (1)

Now, as the box is at rest regarding the vertical direction, the forces acting in this direction (gravity and the normal force) must be equal and opposite each other:

N = Fg = m*g

Replacing in (1) , we have:

Ff = μ*m*g

The work done by the child is as follows:

W = μ*m*g*d

The rate at which the work is done is just the power, which can be expressed as follows:

P = W/t = μ*m*g*(d/t)

but d/t = v (by definition), so we can write the following:

P= μ*m*g*v

0.11 m/s and 0.28 m/s should be correct

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s