Question

A 20.0 kg object is at rest, and then speeds up to 16.0 m/s. How much work was done on this system?

320 J
2,560 J
2,880 J
5,120 J

Answer 1

To find the work done on the object, 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 given by:

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

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

First, we should calculate the initial and final kinetic energies.

1. **Calculate initial kinetic energy (KE_initial):**
Since the object is at rest initially, its initial velocity \(v_i\) is 0 m/s:
\[
KE_{\text{initial}} = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 20.0 \, \text{kg} \times (0 \, \text{m/s})^2 = 0 \, \text{J}
\]

2. **Calculate final kinetic energy (KE_final):**
The final velocity \(v_f\) is 16.0 m/s:
\[
KE_{\text{final}} = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 20.0 \, \text{kg} \times (16.0 \, \text{m/s})^2
\]
\[
KE_{\text{final}} = \frac{1}{2} \times 20.0 \times 256 = 10.0 \times 256 = 2560 \, \text{J}
\]

3. **Calculate the work done (W):**
The work done on the system is equal to the change in kinetic energy:
\[
W = KE_{\text{final}} - KE_{\text{initial}} = 2560 \, \text{J} - 0 \, \text{J} = 2560 \, \text{J}
\]

Thus, the work done on this system is **2,560 J**.