woman is standing on a frozen pond which is essentially frictionless. Next to her is a
large box. The woman’s mass is 45 kg, the box’s mass is 70 kg. The woman pushes the box with
a force such that it is moving away from her at a speed of 2 m/s. (Note: this is the relative
velocity of the box to the woman) How much work did the woman do?

Question

woman is standing on a frozen pond which is essentially frictionless. Next to her is a
large box. The woman’s mass is 45 kg, the box’s mass is 70 kg. The woman pushes the box with
a force such that it is moving away from her at a speed of 2 m/s. (Note: this is the relative
velocity of the box to the woman) How much work did the woman do?

anonymous · Answer

I'm not too sure about this: my answer came out as 453.6 J.
here's what i did:

from the given info, we have Vw - Vb = 2.........(i)
also, the force F acting on both, the woman and the box, will be the same. [from Newton's 3rd law]
so we get,
$$m _{w} a _{w} = m _{b} a _{b}$$ 
$$a _{b} / a _{w} = m _{w} / m _{a}$$
$$V _{b} / V _{w} = m _{w} / m _{b}$$ [using V = u + at = 0 + at = at.....so a = v/t and cancelling out the t's ]

so Vw = Vb mb / mw
sub this in (i) and u'll get the value of Vb as 3.6 m/s.

now, work done = change in KE = 1/2 mb [ (Vb)^2 - (Ub)^2 ] 
W = 0.5 * 70 * (Vb)^2.........since initial speed Ub of box = 0.