When you have pushed the boxes a distance d what is the kinetic energy of box m

Question

When you have pushed the boxes a distance d what is the kinetic energy of box m and which interaction or interactions with objects in its surroundings did the work that produced that final kinetic energy? A box of mass m sits on a larger one of mass M as shown. Both are at rest on a slippery surface when you begin to push as shown on the lower box with a constant force F. the boxes move together (no slipping on each other). its kinetic energy is Fd and the contact force you exerted did the work on box m. Its kinetic energy is m/M + m Fd and the contact force you exerted did the work on box m. its kinetic energy is Fd and the friction force exerted by box M on box m did the work to produce that kinetic energy. its kinetic energy is m/M + m Fd and the friction force exerted by box M on box m did the work to produce that kinetic energy. its kinetic energy is Fd and the force exerted on box M by the floor did the work to produce that kinetic energy.

Explanation / Answer

from the given data

acceleration of both objects, a = F/(M + m) (using Newton's second law)

now frictional force acting on upper block, fs = m*a (using Newton's second law)

= m*F/(M+m)

workdone by friction on the upper block, W = fs*d

= (m*F/(M+m))*d

= (m/(M+m))*F*d

using work-energy theorem,

kinetic energy gained by upper block = workdone on it

= (m/(M+m))*F*d

so, the answer D.

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