Question

A 100 kg box is at rest before a student pushes it across the floor at a velocity of 3 m/s. Calculate the
work being transferred onto the box by the student.
answers:
33.33 J

150 J

450 J

300 J

Answer 1

To find the work being transferred onto the box by the student, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

The formula for kinetic energy (KE) is:

\[
KE = \frac{1}{2}mv^2
\]

where:
- \( m \) is the mass of the object (in kg),
- \( v \) is the velocity of the object (in m/s).

In this scenario:
- The mass \( m = 100 \, \text{kg} \),
- The initial velocity \( v_i = 0 \, \text{m/s} \) (since the box is at rest),
- The final velocity \( v_f = 3 \, \text{m/s} \).

First, we calculate the final kinetic energy:

\[
KE_f = \frac{1}{2} \times 100 \, \text{kg} \times (3 \, \text{m/s})^2
\]

Calculating this:

\[
KE_f = \frac{1}{2} \times 100 \times 9 = 450 \, \text{J}
\]

Since the box was initially at rest, the initial kinetic energy \( KE_i \) is:

\[
KE_i = \frac{1}{2} \times 100 \times (0 \, \text{m/s})^2 = 0 \, \text{J}
\]

Now, we can find the work done (W) on the box:

\[
W = KE_f - KE_i = 450 \, \text{J} - 0 \, \text{J} = 450 \, \text{J}
\]

Thus, the work being transferred onto the box by the student is **450 J**.