A 100 kg box is at rest before a student pushes it across the floor at a velocity of 3 m/s. Calculate the

Question

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:

$$KE=\frac{1}{2}m{v}^2$$

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:

$$K{E}_f=\frac{1}{2}\times 100,\text{kg}\times (3,\text{m/s}{)}^2$$

Calculating this:

$$K{E}_f=\frac{1}{2}\times 100\times 9=450,\text{J}$$

Since the box was initially at rest, the initial kinetic energy $K{E}_i$ is:

$$K{E}_i=\frac{1}{2}\times 100\times (0,\text{m/s}{)}^2=0,\text{J}$$

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

$$W=K{E}_f-K{E}_i=450,\text{J}-0,\text{J}=450,\text{J}$$

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