Power supplied to a particle of mass 2 Kg varies with time as p = 3t^2/2 watt. Here t is in seconds. If initial velocity = 0, find the velocity after 2s.
Sorry for not proving my work. Frankly speaking I'm unable to solve any power questions. Any help would be appreciated. :)
Final ke= Integral power*dt
1/2 m v^2=INT 3t^2/2 dt
= t^3/2 over limits from t=0 to 2
1/2 mv^2=4
v=2 m/s
check that.
No problem! I'll be happy to help you solve this power question step by step.
To find the velocity after 2 seconds, we can use the equation of power:
P = F × v
where P is power, F is force, and v is velocity.
In this case, the power supplied is given by p = 3t^2/2.
Since we are looking for velocity, we need to find the force first. To do that, we can use the equation:
P = F × v
Rearranging the equation:
F = P / v
Plug in the given power, p = 3t^2/2, and rearrange it to solve for force:
F = (3t^2/2) / v
Now, let's consider Newton's second law of motion, which relates force (F) to mass (m) and acceleration (a):
F = m × a
In this case, the force exerted on the particle is the same as the force given by the power equation. Therefore, we can equate them:
(3t^2/2) / v = m × a
The mass of the particle is given as 2 kg, so we have:
(3t^2/2) / v = 2 × a
Now, let's rearrange the equation to solve for acceleration:
a = (3t^2 / 4v)
Since the initial velocity is given as 0, we can use the kinematic equation to find the final velocity:
v = u + at
where u is the initial velocity, a is the acceleration, and t is the time.
Plug in the given values:
u = 0 (initial velocity)
a = (3t^2 / 4v) (acceleration)
t = 2 seconds (time)
v = 0 + [(3(2)^2) / (4v)] × 2
Simplifying further:
v = 2v
Now, solve the equation:
v = 2v
Divide both sides by 2:
v / 2 = v
Subtract v from both sides:
v - v / 2 = 0
Simplify further:
v/2 = 0
Multiply both sides by 2:
v = 0
Therefore, the velocity after 2 seconds is 0 m/s.