Explanation
Given,
x=3t−4t2+t3
dx=d(3t−4t2+t3)=(3−8t+3t2)dt
Acceleration
a(x)=d2xdt=d2xdt=d(3−8t+3t2)dt=−8+6t
Work done =dw=ma.dx=m(6t−8)(3−8t+3t2)dt
∫W0dw=m∫40(6t−8)(3−8t+3t2)dt
W=m×176=0.003×176=528×10−3J
W=528×10−3J
Final velocity, v1
Change in kinetic Energy = work done by force
12mv21−12mv2=W
v1=√2Wm+v2
Hence, final velocity is √2Wm+v2
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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