Work Energy and Power Test - 46

$$v=a\sqrt{x}\quad \Rightarrow \cfrac{dx}{dt}=ax^{1/2}\quad \Rightarrow \int x^{1/2} dx=\int a dt$$
$$2\sqrt{x}=at\quad \Rightarrow x=\cfrac{a^2t^2}{4}$$
$$\Rightarrow v=\cfrac{a^2t}{2}$$
$$\Rightarrow a^{\prime}=\cfrac{a^2}{2}$$
$$F=ma^{\prime}=\cfrac{ma^2}{2}$$
$$Work = Force\times Displacement$$
$$\Rightarrow Work =\cfrac{ma^2}{2}\times 4=2ma^2$$
As, $$m=2$$
$$\therefore Work = 4a^2$$

 It is given that,

 $$ F=x\hat{i}+2y\hat{j} $$

 $$ Let,\,\,dx=dx\hat{i}+dy\hat{j} $$

 $$ dw=\int{F.dx} $$

 $$ =\int{(x\hat{i}+2y\hat{j}})(dx\hat{i}+dy\hat{j}) $$

 $$ =\int\limits_{1}^{0}{xdx}+2\int\limits_{2}^{1}{ydy} $$

 $$ ={{\left. \dfrac{{{x}^{2}}}{2} \right|}_{1}}^{0}+{{\left. {{y}^{2}} \right|}_{2}}^{1} $$

 $$ =\dfrac{-1}{2}+(1-4) $$

 $$ =-0.5-3 $$

 $$ =-3.5\,\,J $$

 

$$ T - mg = \cfrac{mv^2}{r}$$

Given that,

Force $$F=KV$$

Velocity=$$V$$

We know that,

$$F=Ma$$

Now,

  $$ Ma=KV $$

 $$ M\left( \dfrac{dV}{dt} \right)=KV $$

 $$ MdV=KVdt $$

Multiply by v in both side

$$MVdV=K{{V}^{2}}dt$$

On integrating both side

  $$ \int\limits_{{{v}_{1}}}^{{{v}_{2}}}{MVdV=\int\limits_{0}^{t}{K{{V}^{2}}dt}} $$

 $$ M\left[ \dfrac{V_{2}^{2}}{2}-\dfrac{V_{1}^{2}}{2} \right]=K{{V}^{2}}t $$

 $$ \dfrac{M}{2}\left[ V_{2}^{2}-V_{1}^{2} \right]=K{{V}^{2}}t $$

Now, the change in K.E

Now, from work energy theorem

Change in energy = work done

$$K.E=K{{V}^{2}}t$$

Hence, the work done is $$K{{V}^{2}}t$$ 

Given that,

Mass $$m=2\,kg$$  

Distance $$x=\dfrac{{{t}^{2}}}{4}$$

Now, on differentiate

$$\dfrac{dx}{dt}=\dfrac{1}{2}t$$

Now, velocity and acceleration is

$$ v=\dfrac{dx}{dt}=\dfrac{t}{2} $$

 $$ a=\dfrac{dv}{dt}=\dfrac{1}{2} $$

Now, the force is

 $$ F=ma $$

 $$ F=2\times \dfrac{1}{2} $$

 $$ F=1\,N $$

Now, the work done is

  $$ dW=F\centerdot dx $$

 $$ \int{dW}=\int\limits_{0}^{2}{1}\times \dfrac{t}{2}dt $$

 $$ W=\left[ \dfrac{{{t}^{2}}}{4} \right]_{0}^{2} $$

 $$ W=\left[ \dfrac{4}{4}-0 \right] $$

 $$ W=1\,J $$

Hence, the work done is $$1\ J$$