(a) A 3.0-g locust reaches a speed of \({\bf{3}}{\bf{.0}}\;{{\bf{m}} \mathord{\left/{\vphantom {{\bf{m}} {\bf{s}}}} \right.} {\bf{s}}}\)during its jump. What is its kinetic energy at this speed? (b) If the locust transforms energy with 35% efficiency, how much energy is required for the jump?

The kinetic energy is the particle's energy due to its state of movement. It can be shifted from one particle to another.

The mass of the locust is\(m = 3.0\;{\rm{g}}\).

The speed of the locust is \(v = 3.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).

(a)

The kinetic energy of the locust can be calculated as shown below:

\(\begin{aligned}K{E_{\rm{a}}} &= \frac{1}{2}m{v^2}\\K{E_{\rm{a}}} &= \frac{1}{2}\left( {\left( {{\rm{3}}{\rm{.0}}\;{\rm{g}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{kg}}}}{{{\rm{1}}\;{\rm{g}}}}} \right)} \right){\left( {3.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)^2}\\K{E_{\rm{a}}} &= 1.35 \times {10^{ - 2}}\;{\rm{J}}\end{aligned}\)

Thus, the kinetic energy of the locust is \(1.35 \times {10^{ - 2}}\;{\rm{J}}\).

(b)

The locust transforms energy with an efficiency of 35%.

The relation between the actual and required energy is\(K{E_{\rm{a}}} = \left( {0.35} \right){E_{\rm{r}}}\).

It can be rewritten as

\({E_{\rm{r}}} = \frac{{K{E_a}}}{{0.35}}\).

Substitute the values in the above equation.

\(\begin{aligned}{E_{\rm{r}}} &= \frac{{\left( {1.35 \times {{10}^{ - 2}}\;{\rm{J}}} \right)}}{{\left( {0.35} \right)}}\\{E_{\rm{r}}} &= 0.039\;{\rm{J}}\end{aligned}\)

Thus, the energy required for the jump is \(0.039\;{\rm{J}}\).