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.