Question 337: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

To determine the speed of a particle after $ t = 5 $ seconds, given a constant power, the following steps should be taken :
Power Equation:
Power is the rate at which work is done, and work is the change in kinetic energy :
$$ P = \frac{d(W)}{dt} = \frac{d}{dt} \left( \frac{1}{2} m v^2 ight) $$
Given Data :
Mass of the particle ($ m $) = 0.2 kg
Constant power ($ P $) = 0.5 W
Initial speed ($ v_0 $) = 0 m/s
Differentiating Kinetic Energy with Respect to Time :
$$ P = \frac{d}{dt} \left( \frac{1}{2} m v^2 ight) = m v \frac{dv}{dt} $$
Rearranging to Solve for $ v \frac{dv}{dt} $ :
$$ 0.5 = 0.2 \cdot v \cdot \frac{dv}{dt} $$
$$ \frac{dv}{dt} = \frac{5}{2v} $$
Integrating Both Sides :
$$ v \, dv = \frac{5}{2} \, dt $$
$$ \int v \, dv = \int \frac{5}{2} \, dt $$
$$ \frac{v^2}{2} = \frac{5t}{2} + C $$
Applying Initial Condition (when $ t = 0 $, $ v = 0 $) :
$$ 0 = \frac{5 \cdot 0}{2} + C $$
$$ C = 0 $$
Solving for $ v $ :
$$ \frac{v^2}{2} = \frac{5t}{2} $$
$$ v^2 = 5t $$
$$ v = \sqrt{5t} $$
Speed After 5 Seconds :
$$ v = \sqrt{5 \cdot 5} = \sqrt{25} = 5 \, 	ext{m/s} $$
Thus, the speed of the particle after 5 seconds is
5 m/s
.

Work, Energy and Power & Circular Motion

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

Integer / Numeric Answer Type