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Question

Physics Question on Conservation of energy

Potential energy of a body of mass 1kg1\, kg free to move along X-axis is given by U(x)=(x22x)J.U(x) = \left( \frac{x^2}{2} - x \right) J . If the total mechanical energy of the body is 2J2\, J. then the maximum speed of the body is (Assume only conservative force acts on the body)

A

5ms1\sqrt{5}\, ms^{-1}

B

5  ms15 \; ms^{-1}

C

3.5  ms13.5 \; ms^{-1}

D

8  ms1\sqrt{8} \; ms^{-1}

Answer

5ms1\sqrt{5}\, ms^{-1}

Explanation

Solution

Total mechanical energy of a system is the addition of potential and kinetic energy
E=U+KEE=U+ KE
Here, mass of the body, m=1kgm=1 \,kg,
Emech =2JE_{\text {mech }}=2 \,J
So, for Umin=dU(x)dx=ddx[x22x]=0U_{\min }=\frac{d U(x)}{d x}=\frac{d}{d x}\left[\frac{x^{2}}{2}-x\right]=0
x1=0\Rightarrow x - 1 = 0
x=1\Rightarrow x = 1
Hence, Umin=(1)221=12U_{\min }=\frac{(1)^{2}}{2}-1=-\frac{1}{2}
Kinetic energy. KF=12mv2=v22(m=1kg)KF =\frac{1}{2} m v^{2}=\frac{v^{2}}{2} \,\,(\because m=1 \,kg )
Now, putting the values in above expression,
2=12+v22\Rightarrow 2=-\frac{1}{2}+\frac{v^{2}}{2}
v2=5\Rightarrow v^{2}=5
v=5m/s\Rightarrow v=\sqrt{5}\, m / s