A box slides along a rough floor with initial kinetic energy of 7 J. A person pushes on the box, in the same direction as its velocity doing 25 J of work. The velocity of the box is doubled. Find the magnitude of the work done by friction.
initial KE=7
final KE=7+25-frictionwork
velocity is doubled, so KE is quadrupled.
final KE=28J
28=7+25-friction
solve for friction
To find the magnitude of the work done by friction, we need to consider the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy. In this case, we know the initial kinetic energy of the box (7 J) and the work done by the person who pushes the box (25 J).
Let's break down the different contributors to the net work done on the box:
1. The work done by the person pushing the box (positive work): 25 J
2. The work done by friction (negative work): ? (We need to find this)
3. The net work done on the box: change in kinetic energy
Since the velocity of the box is doubled, we can calculate the change in kinetic energy as follows:
Change in kinetic energy = (1/2) * m * (v^2 - u^2)
where m is the mass of the box, and u and v are the initial and final velocities, respectively.
Since the velocity is doubled, the final velocity is 2u (where u is the initial velocity).
Change in kinetic energy = (1/2) * m * ((2u)^2 - u^2)
= (1/2) * m * (4u^2 - u^2)
= (1/2) * m * 3u^2
Given that the initial kinetic energy is 7 J, we have:
7 J = (1/2) * m * 3u^2
Now we can solve for the mass of the box (m):
m = (7 J) * 2 / (3u^2)
With the mass (m) and the velocity (u) known, we can now calculate the work done by friction:
Work done by friction = - change in kinetic energy - work done by the person
= - (7 J - 0) - 25 J
= - 7 J - 25 J
= - 32 J
Therefore, the magnitude of the work done by friction is 32 J.