CBSE Questions for Class 11 Medical Physics Work, Energy And Power Quiz 8 - MCQExams.com

Given,

Final velocity, $${{v}_{1}}$$

Change in kinetic Energy = work done by force

$$ \dfrac{1}{2}mv_{1}^{2}-\dfrac{1}{2}m{{v}^{2}}=W $$

$$ {{v}_{1}}=\sqrt{\dfrac{2W}{m}+{{v}^{2}}} $$

Hence, final velocity is $$\sqrt{\dfrac{2W}{m}+{{v}^{2}}}$$

 It is given that,

 $$ F=x\hat{i}+2y\hat{j} $$

 $$ Let,\,\,dx=dx\hat{i}+dy\hat{j} $$

 $$ dw=\int{F.dx} $$

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

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

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

 $$ =\dfrac{-1}{2}+(1-4) $$

 $$ =-0.5-3 $$

 $$ =-3.5\,\,J $$

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 $$1\ J$$

Given that,

Mass of body $$m=10\,kg$$

Height from depth $$h=10\,m$$

Velocity $$v=2\,m/s$$

Now,

Work done =   work done to change in P.E + work done in providing K.E to body

 $$ W=mgh+\dfrac{1}{2}m{{v}^{2}} $$

 $$ W=10\times 9.8\times 10+\dfrac{1}{2}\times 10\times 2\times 2 $$

 $$ W=980+20 $$

 $$ W=1000\,J $$

Hence, the work done is $$1000\ J$$

Given,

Force, $$F=\left( 3t\hat{i}+5\hat{j} \right)N$$

Displacement, $$s=\left( 2{{t}^{2}}\hat{i}-5\hat{j} \right)\,m$$

 $$ds=\left( 4t\hat{i}+0\hat{J} \right)\,dt$$

 $$ dw=F.ds $$

 $$ \int_{0}^{w}{dw}=\int_{0}^{2}{\left( 3t\hat{i}+5\hat{j} \right)}\left( 4t\hat{i}+0\hat{j} \right)\,dt $$

 $$ W=\left. 12{{t}^{2}} \right|_{0}^{2}=48\,J $$

Net work done is $$48\,J$$ 

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,

Velocity, $$\dfrac{dx}{dt}=v=k{{x}^{3/2}}$$

Acceleration,

$$ a=\dfrac{dv}{dt}=\dfrac{d\left( k{{x}^{3/2}} \right)}{dt}=\dfrac{1}{2}k{{x}^{1/2}}\dfrac{dx}{dt}=\dfrac{1}{2}k{{x}^{1/2}}v $$

$$ a=\dfrac{1}{2}k{{x}^{1/2}}\left( k{{x}^{3/2}} \right)=\dfrac{1}{2}{{k}^{2}}{{x}^{2}} $$

$$ work\,done\,=\,force\times displacemet\,=Mass\times acceleration\times displacement $$

 $$ \Rightarrow dW=m.a\,.dx=m.\dfrac{1}{2}{{k}^{2}}{{x}^{2}}dx $$

On integrating

$$W=m\dfrac{1}{2}{{k}^{2}}\left( \dfrac{{{x}^{3}}}{3} \right)$$

Hence, Work done is proportional to $${{x}^{3}}$$ 

Spring energy $$E=\dfrac{1}{2}k{{x}^{2}}=\dfrac{1}{2}80\times {{0.12}^{2}}=0.576\,J$$

From conservation of energy

Maximum Spring energy = Maximum kinetic energy of ball

$$ E=\dfrac{1}{2}M{{V}^{2}} $$

$$ V=\dfrac{2E}{M}=\dfrac{2\times 0.576}{0.015}=76.8m{{s}^{-1}} $$

Velocity of ball is $$76.8\,m{{s}^{-1}}$$