Question

Particle of mass m moves under the influence of force f =a( sinwt i^ +coswt j^) where a ,w are constants and t is time . the particle is initially at rest atthe origin . the instantaneous power given to the particle is

Answer 1

Given that,

Mass = m

Time = t

Force $f=a(\sin\omega t\hat{i}+\cos\omega t\hat{j})$

We need to calculate the acceleration

Using formula of force

$F= ma$

$a=\dfrac{F}{m}$

$a=\dfrac{a(\sin\omega t\hat{i}+\cos\omega t\hat{j})}{m}$

We need to calculate the velocity

Using formula of velocity

$v=\int{a}dt$

$v=\int{\dfrac{a(\sin\omega t\hat{i}+\cos\omega t\hat{j})}{m}}dt$

$v=\dfrac{a}{m}(-\dfrac{1}{\omega}\cos\omega t\hat{i}+\dfrac{1}{\omega}\sin\omega t\hat{j})+C$

At t = 0, v = 0

Then the value of C will be

$0=-\dfrac{a}{m\omega}\hat{i}+C$

$C=\dfrac{a}{m\omega}\hat{i}$

Put the value of C

$v=\dfrac{a}{m}(-\dfrac{1}{\omega}\cos\omega t\hat{i}+\dfrac{1}{\omega}\sin\omega t\hat{j})+\dfrac{a}{m\omega}\hat{i}$

$v=\dfrac{a}{\omega m}(1-\cos\omega t)\hat{i}+\sin\omega t\hat{j}$

We need to calculate the work

Using formula of work

$W=\int_{0}^{t}{F\cdot v dt}$

$W=\int_{0}^{t}{a(\sin\omega t\hat{i}+\cos\omega t\hat{j})\cdot\dfrac{a}{\omega m}((1-\cos\omega t)\hat{i}+\sin\omega t\hat{j})}dt$

$W=\dfrac{a^2}{\omega m}\int_{0}^{t}{(\sin\omega t-\sin\omega t\cos\omega t+\sin\omega t\cos\omega t})dt$

$W=\dfrac{a^2}{\omega m}(-\dfrac{1}{\omega}\cos\omega t)_{0}^{t}$

$W=\dfrac{a^2}{\omega^{2}m}(1-\cos\omega t)$

We need to calculate the power

Using formula of power

$P=\dfrac{dW}{dt}$

Put the value in to the formula

$P=\dfrac{a^2}{\omega^{2}m}(0+\omega\sin\omega t)$

$P=\dfrac{a^2}{\omega m}\sin\omega t$

Hence, The instantaneous power given to the particle is $\dfrac{a^2}{\omega m}\sin\omega t$

Answer 2

Explanation:

The solution is attached herewith.