Question

a 0.6 kg object moves at a constant velocity of 2 m/s, traveling a total distance of 30 m. using the work - energy theorem, what is the net work on the object? (1 point)

1.2 j

0 j

36 j

2.4 j

Explanation:

Step1: Recall work - energy theorem

The work - energy theorem is $W_{net}=\Delta K$, where $W_{net}$ is the net work done on the object and $\Delta K$ is the change in kinetic energy.

Step2: Calculate initial and final kinetic energies

The kinetic energy formula is $K = \frac{1}{2}mv^{2}$. Since the object moves at a constant velocity $v = 2\ m/s$, the initial kinetic energy $K_{i}=\frac{1}{2}(0.6\ kg)(2\ m/s)^{2}=\frac{1}{2}\times0.6\times4\ J = 1.2\ J$, and the final kinetic energy $K_{f}=\frac{1}{2}(0.6\ kg)(2\ m/s)^{2}=1.2\ J$.

Step3: Find the change in kinetic energy

$\Delta K=K_{f}-K_{i}=1.2\ J - 1.2\ J=0\ J$.

Step4: Determine the net work

Since $W_{net}=\Delta K$, then $W_{net}=0\ J$.

Answer:

0 J