Question

A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power of 0.5 W to the particle. If the initial speed (in m/s) of the particle is zero, the speed (in m/sec) after 5 s is?

Answer 1

In this given problem power delivered by the moving particle under the influence of a force is constant.
And we know that power can be defined as the rate of doing work.
And work is said to be done by a force when the body on which force is applied, is displaced by some distance in the direction of applied force.

Complete step by step answer:
Step 1:
As we know from the definition of work done that it is defined as –
Work done = force x displacement
$W = F \times S$ …………..(1)
And force can be defined as –
Force = Mass x acceleration
$F = M \times a$ …………..(2)
Where $M$is mass of the particle, and $a$ is acceleration of the particle
So, from equation (1) and (2)
$W = MaS$ …………..(3)
Multiply and divide the equation with 2 and rearranging it, we will get –
$W = \dfrac{2}{2}MaS = \dfrac{M}{2}2aS$ …………..(4)

Step 2: We know the third equation of motion and from the third equation of motion the value of $2aS$ can be calculated as the equation given below –
$\mathop v\nolimits^2 = \mathop u\nolimits^2 + 2aS$ …………..(5)
Where $v$ is the final velocity of the particle, $u$ is the initial velocity of the particle, $a$is acceleration of the particle, and $S$ is displacement
On rearranging the equation (5)
$2aS = \mathop v\nolimits^2 - \mathop u\nolimits^2$ but in the question, it is given that $u$ is 0m/s
So $2aS = \mathop v\nolimits^2$ …………..(6)
Substituting the value from equation (6) into equation (5), we will get –
$W = \dfrac{M}{2}\mathop v\nolimits^2$ …………..(7)

Step 3: But power at particular instant of time can be defined as
Power = small work done / small time interval
i.e., $P = \dfrac{{dW}}{{dt}}$ After rearranging this equation for the time interval from 0s to 5s, we will get –
$\int {dW} = \int\limits_0^5 {Pdt}$ and $W = \int\limits_0^5 {Pdt}$ …………..(8)
Form equation (7) and (8)
$\dfrac{M}{2}\mathop v\nolimits^2 = \int\limits_0^5 {0.5dt}$
On rearranging the above equation and solving the integration, we will get –
$\mathop v\nolimits^2 = \dfrac{2}{M} \times 0.5\mathop {\left[ t \right]}\nolimits_0^5$ after substituting the given values in this equation
$\mathop v\nolimits^2 = \dfrac{2}{{0.2}} \times 0.5\left[ {5 - 0} \right]$
$\mathop v\nolimits^2 = 25$m/s
$v = 5$m/s.
So, the final velocity will be 5m/s.

$\therefore$ The speed after 5s is 5m/s.

Note:
The given problem is an example of constant force so work is done is calculated by the given formula but if the force is a variable force working on the body then work done can be calculated by the area under the force and displacement curve.