Question

Velocity of a particle of mass $1\,{\text{kg}}$ moving rectilinearly is given by $v = 25 - 2t + {t^2}$. Find the average power of the force acting on the particle between time intervals $t = 0$ to $t = 1\,{\text{s}}$.
A. $49\,{\text{W}}$
B. $24.5\,{\text{W}}$
C. $- 49\,{\text{W}}$
D. $- 24.5\,{\text{W}}$

Answer 1

Use the formula for kinetic energy of the particle in terms of its mass and velocity. Use the expression for work-energy theorem to determine the work done by the particle. Use the formula for power in terms of work and time to determine the power of the force acting on the particle.

Formulae used:
The kinetic energy $K$ of an object is
$K = \dfrac{1}{2}m{v^2}$ …… (1)
Here, $m$ is the mass of the object and $v$ is the velocity of the object.
The expression for work-energy theorem is
$W = \Delta K$ …… (2)
Here, $W$ is the work done and $\Delta K$ is a change in the kinetic energy.
The power $P$ is given by
$P = \dfrac{W}{t}$ …… (3)
Here, $W$ is the work done and $t$ is the time.

Complete step by step answer:
We have given that the mass of the particle is $1\,{\text{kg}}$..
$m = 1\,{\text{kg}}$

The velocity of the particle is given by
$v = 25 - 2t + {t^2}$ …… (4)
We can determine the velocities at times $t = 0\,{\text{s}}$ and $t = 1\,{\text{s}}$ from the above equation.

Substitute $0\,{\text{s}}$ for $t$ in equation (4).
${v_i} = 25 - 2\left( {0\,{\text{s}}} \right) + {\left( {0\,{\text{s}}} \right)^2}$
$\Rightarrow {v_i} = 25\,{\text{m/s}}$
Hence, the initial velocity of the particle is $25\,{\text{m/s}}$.

Substitute $1\,{\text{s}}$ for $t$ in equation (4).
${v_f} = 25 - 2\left( {1\,{\text{s}}} \right) + {\left( {1\,{\text{s}}} \right)^2}$
$\Rightarrow {v_f} = 24\,{\text{m/s}}$
Hence, the final velocity of the particle is $24\,{\text{m/s}}$.

The change in kinetic energy of the particle is equal to the difference of the final and initial kinetic energy of the particle.
$\Delta K = {K_f} - {K_i}$
Hence, equation (2) becomes
$W = {K_f} - {K_i}$

Substitute the values of initial and final kinetic energies in the above equation using equation (1).
$W = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2$

Substitute $1\,{\text{kg}}$ for $m$, $25\,{\text{m/s}}$ for ${v_i}$ and $24\,{\text{m/s}}$ for ${v_f}$ in the above equation.
$W = \dfrac{1}{2}\left( {1\,{\text{kg}}} \right){\left( {24\,{\text{m/s}}} \right)^2} - \dfrac{1}{2}\left( {1\,{\text{kg}}} \right){\left( {25\,{\text{m/s}}} \right)^2}$
$\Rightarrow W = \dfrac{1}{2}\left( {576 - 625} \right)$
$\Rightarrow W = - 24.5{\text{ J}}$
Hence, the work done by the particle is $- 24.5{\text{ J}}$.

Let us determine the power of the force acting on the particle in a time interval of 1 second using equation (3).
Substitute $- 24.5{\text{ J}}$ for $W$ and $1\,{\text{s}}$ for $t$ in equation (3).
$P = \dfrac{{ - 24.5{\text{ J}}}}{{1\,{\text{s}}}}$
$\therefore P = - 24.5{\text{ W}}$

Therefore, the power of the force acting on the particle is $- 24.5{\text{ W}}$. Hence, the correct option is D.

Note: The work done and the power due to the force acting on the particle are negative because the velocity of the particle determined by the given relation of velocity with time is decreasing. Hence, the work done by the particle is also negative resulting in negative power in the particular time interval.