Explanation
Given,
$$ x=3t-4{{t}^{2}}+{{t}^{3}} $$
$$ dx=d\left( 3t-4{{t}^{2}}+{{t}^{3}} \right)=\left( 3-8t+3{{t}^{2}} \right)dt $$
Acceleration
$$a(x)=\dfrac{{{d}^{2}}x}{dt}=\dfrac{{{d}^{2}}x}{dt}=\dfrac{d\left( 3-8t+3{{t}^{2}} \right)}{dt}=-8+6t$$
Work done =$$dw=ma.dx=m\left( 6t-8 \right)\left( 3-8t+3{{t}^{2}} \right)dt$$
$$ \int_{0}^{W}{dw}=m\int_{0}^{4}{\left( 6t-8 \right)\left( 3-8t+3{{t}^{2}} \right)dt} $$
$$ W=m\times 176=0.003\times 176=528\times {{10}^{-3}}\,J $$
$$ W=528\times {{10}^{-3}} J $$
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 $$\sqrt{\dfrac{2W}{m}+{{v}^{2}}}$$
We know that, the kinetic energy is
$$K.E=\dfrac{1}{2}m{{v}^{2}}....(I)$$
Checking all options given -
A) $$ m=3M $$ & $$ v=v $$
$$ K.E=\dfrac{1}{2}m{{v}^{2}} $$
$$ K.E=\dfrac{1}{2}\times 3M\times {{v}^{2}} $$
$$ K.E=\dfrac{3M{{v}^{2}}}{2} $$
B) $$ m=3M $$ & $$ v=2v $$
$$ K.E=\dfrac{1}{2}\times 3M\times 4{{v}^{2}} $$
$$ K.E=6M{{v}^{2}} $$
C) $$ m=2M $$ & $$ v=3v $$
$$ K.E=\dfrac{1}{2}\times 2M\times 9{{v}^{2}} $$
$$ K.E=9M{{v}^{2}} $$
D) $$ m=M $$ & $$ v=4v $$
$$ K.E=\dfrac{1}{2}\times M\times 16{{v}^{2}} $$
$$ K.E=8M{{v}^{2}} $$
Hence, the largest kinetic energy is $$9M{{v}^{2}}$$ when $$m=2M$$ and $$v=3v$$.
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 $$48\,J$$
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,
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 $$K{{V}^{2}}t$$
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 $$1\ J$$
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 $${{x}^{3}}$$
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