Q: A particle in a certain conservative force field has a potential energy given by \$U=\dfrac{20xy}{z}\$. The force exerted on it is?

The potential energy function \$U(x)\$ for a system in which a one - dimensional force F acts on a particle, we can find the force as \$F(x)=\dfrac { dU }{ dx } =\dfrac { dU_{ x } }{ dx } +\dfrac { dU_{ y } }{ dx } +\dfrac { dU_{ z } }{ dx } \$ Apply \$U=20\dfrac{xy}{z}\$\$=\dfrac { 20y }{ z } i+\dfrac { 20x }{ z } j−\dfrac { 20xy }{ z^2 }k \$

Q: Calculate the work done in lifting water of 200 kg through a vertical distance of 6 m. \$\left ( g=10 m s^{-2} \right )\$

Work done in lifting water = \$mgh\$ \$=200\times 10\times 6 = 12000\ J\$

Q: A chain of mass \$M\$ is placed on a smooth table with \$1/n\$ of its length \$L\$ hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is:

Mass per unit length \$=\dfrac{M}{L}\$mass of dx element \$=\dfrac{M}{L}dx\$ \$dm=\dfrac{M}{L}dx\$Work done by pulling's dx element over the table \$=dw=(dm \quad x)g\$\$dw=\dfrac{Mg}{L} xdx\$\$w=\int _{ 0 }^{ \left( \dfrac { L }{ n } \right) }{ \dfrac { Mg }{ L } } x dx\\ =\dfrac { Mg }{ L } \overset { \dfrac { L }{ n } }{ \underset { 0 }{ \left[ \dfrac { { x }^{ 2 } }{ 2 } \right] } } \\ =\dfrac { Mg }{ L } \dfrac { { L }^{ 2 } }{ 2{ n }^{ 2 } } =\dfrac { mgL }{ 2{ n }^{ 2 } } \$

Q: The diagram given below shows a ski jump. A skier weighing \$60\, kgf\$ stands at A at the top of ski jump. He moves from A to B and takes off for his jump at B. (a) Calculate the change in the gravitational potential energy of the skier between A and B. (b) If \$75\$% of the energy in part (a) becomes kinetic energy at B. Calculate the speed at which the skier arrives at B. \$(Take \,g=10\, m\, s^{-1})\$

(a) Mass of skier \$= 60\,kg\$ Loss in potential energy \$= mg(h_1 -h_2)\$\$ 60 \times 10 \times (75 -15) \$\$= 60 \times 10 \times 60 = 3.6 \times 10^4\, J\$ (b) Kinetic energy at \$B = \dfrac{75}{100} \times 3.6 \times 70^4 = 27000 \, J\$\$ = 2.7 \times 10^4\, J\$ Kinetic energy \$=\dfrac{1}{2} mv^2\$ \$27000 =\dfrac {1}{2} mv^2\$ \$27000 =\dfrac {1}{2} \times 60 \times v^2\$\$v^2 =\dfrac {27000}{30} = 900\$\$V= 30/, m/s\$

Q: A chain of length \$l\$ and mass \$m\$ lies of the surface of a smooth hemisphere of radius \$R>1\$ with one end tied to the top of the hemisphere. Taking base of the hemisphere as reference line, find the gravitational potential energy of the chain.

Given, Mass of chain is \$m\$ Length is \$l\$ Radius of sphere is \$R\$ \$\Rightarrow h=Rsin\theta\$ \$\Rightarrow dl=Rd\theta\$ The mass of dl length of the chain \$dm=mdl\$ Potential energy. \$d U = ( d m ) g h\$ \$\Rightarrow dU=\dfrac{mdl}{l}gh=\dfrac{mghdl}{l}=\dfrac{mg{{h}^{2}}\sin \theta d\theta }{l}\$ \$\Rightarrow dU=\dfrac{mg{{R}^{2}}\sin \theta d\theta }{l}\$ On Integrating \$U=\int_{\theta }^{\dfrac{x}{2}}{\dfrac{mg{{R}^{2}}\sin \theta d\theta }{l}}\$ \$U=-\dfrac{mg{{R}^{2}}}{l}[\cos \theta ]_{\theta }^{\dfrac{\pi }{2}}=-\dfrac{mg{{R}^{2}}}{l}\left[ \cos \left( \dfrac{\pi }{2} \right)-\cos \theta \right]\$ \$U=-\dfrac{mg{{R}^{2}}}{l}\cos \theta =\dfrac{mg{{R}^{2}}}{l}\sin \left( \dfrac{l}{R} \right)\$ Hence, potential energy is \$\dfrac { m g R ^ { 2 } } { l } \sin \left( \dfrac { l } { R } \right)\$

Q: Water stored in a dam possesses:

The potential energy possessed by the object is the energy present in it by virtue of its position or configuration. Water stored in a dam , when allowed to flow, has kinetic energy which was earlier stored as potential energy.

Q: A body of mass 0.2 kg falls from a height of 10m to a height of 6m above the ground. Find the loss in potential energy taking place in the body. [\$g=10ms^{-2}\$].

Given : \$m= 0.2 kg, g= 10 ms^{-2}, h_1=10m, h^2=6m\$.Loss in P.E. \$=mg(h_1-h_2)\$\$= 0.2 \times 10 \times 10(10-6)\$\$=0.2 \times 10 \times 4=8 J\$

Question 1

A particle in a certain conservative force field has a potential energy given by $U=\dfrac{20xy}{z}$. The force exerted on it is?

Accepted Answer

The potential energy function $U(x)$ for a system in which a one - dimensional force F acts on a particle, we can find the force as&#10;  $F(x)=\dfrac { dU }{ dx } =\dfrac { dU_{ x } }{ dx } +\dfrac { dU_{ y } }{ dx } +\dfrac { dU_{ z } }{ dx } $         Apply $U=20\dfrac{xy}{z}$$=\dfrac { 20y }{ z } i+\dfrac { 20x }{ z } j&#8722;\dfrac { 20xy }{ z^2 }k $

Question 2

A body is dropped from a certain height from the ground. When it is halfway down, it possess,

Accepted Answer

when a body is at some height from ground then its potential energy is $mgh$, assuming ground surface as reference.also when the body has moved some distance than some of potential energy is converted to kinetic energy so it has both the energies kinetic as well as potential.

Question 3

A person pulls a bucket of water from a well of depth $h$. If the mass of uniform rope is $m$ and that of the bucket full of water is $M$, then work done by the person is:

Accepted Answer

Work done by the person is the change in potential energy of the bucket and the rope.Given, the rope has a uniform mass $m$. So the center of mass for the rope is at half of the length of the rope.So, it will be at a distance $h/2$ from the bottom.Work done = Change in PE of bucket+Change in PE of rope$W=Mgh+mg\dfrac{h}{2}=(M+\dfrac{m}{2})gh$Option A.

Question 4

Calculate the work done in lifting  water of 200 kg through a vertical distance of 6 m. $\left ( g=10  m  s^{-2} \right )$

Accepted Answer

Work done in lifting water = $mgh$                                         $=200\times 10\times 6 = 12000\ J$

Question 5

A chain of mass $M$ is placed on a smooth table with $1/n$ of its length $L$ hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is:

Accepted Answer

Mass per unit length $=\dfrac{M}{L}$mass of dx element  $=\dfrac{M}{L}dx$ $dm=\dfrac{M}{L}dx$Work done by pulling's dx element over the table $=dw=(dm \quad x)g$$dw=\dfrac{Mg}{L} xdx$$w=\int _{ 0 }^{ \left( \dfrac { L }{ n }  \right)  }{ \dfrac { Mg }{ L }  }  x dx\ =\dfrac { Mg }{ L } \overset { \dfrac { L }{ n }  }{ \underset { 0 }{ \left[ \dfrac { { x }^{ 2 } }{ 2 }  \right]  }  } \ =\dfrac { Mg }{ L } \dfrac { { L }^{ 2 } }{ 2{ n }^{ 2 } } =\dfrac { mgL }{ 2{ n }^{ 2 } } $

Question 6

The diagram given below shows a ski jump. A skier weighing $60 k g f$ stands at A at the top of ski jump. He moves from A to B and takes off for his jump at B.
(a) Calculate the change in the gravitational potential energy of the skier between A and B.
(b) If $75$ % of the energy in part (a) becomes kinetic energy at B. Calculate the speed at which the skier arrives at B.
$(T a k e g = 10 m s^{- 1})$

Accepted Answer

(a) Mass of skier $= 60\,kg$ Loss in potential energy $= mg(h_1 -h_2)$$ 60 \times 10  \times (75 -15) $$= 60  \times 10  \times 60 = 3.6  \times 10^4\, J$ (b) Kinetic energy at $B = \dfrac{75}{100}  \times 3.6  \times 70^4 = 27000 \, J$$ = 2.7  \times 10^4\, J$ Kinetic energy $=\dfrac{1}{2} mv^2$ $27000 =\dfrac {1}{2} mv^2$ $27000 =\dfrac {1}{2}   \times 60  \times v^2$$v^2 =\dfrac {27000}{30} = 900$$V= 30/, m/s$

Question 7

A chain of length $l$ and mass $m$ lies of the surface of a smooth hemisphere of radius $R>1$ with one end tied to the top of the hemisphere. Taking base of the hemisphere as reference line, find the gravitational potential energy of the chain.

Accepted Answer

Given,&#10;&#10;Mass of&#10;chain is $m$&#10;&#10;Length&#10;is $l$&#10;&#10;Radius&#10;of sphere is $R$&#10;&#10;$\Rightarrow&#10;h=Rsin\theta$&#10;&#10;$\Rightarrow&#10;dl=Rd\theta$&#10;&#10;The&#10;mass of dl length of the chain $dm=mdl$&#10;&#10;Potential&#10;energy.&#10;&#10;$d U =&#10;( d m ) g h$&#10;&#10;$\Rightarrow&#10;dU=\dfrac{mdl}{l}gh=\dfrac{mghdl}{l}=\dfrac{mg{{h}^{2}}\sin \theta d\theta }{l}$&#10;&#10;$\Rightarrow&#10;dU=\dfrac{mg{{R}^{2}}\sin \theta d\theta }{l}$&#10;&#10;On&#10;Integrating&#10;&#10;$U=\int_{\theta&#10;}^{\dfrac{x}{2}}{\dfrac{mg{{R}^{2}}\sin \theta d\theta }{l}}$&#10;&#10;$U=-\dfrac{mg{{R}^{2}}}{l}[\cos&#10;\theta ]_{\theta }^{\dfrac{\pi }{2}}=-\dfrac{mg{{R}^{2}}}{l}\left[ \cos \left(&#10;\dfrac{\pi }{2} \right)-\cos \theta  \right]$&#10;&#10;$U=-\dfrac{mg{{R}^{2}}}{l}\cos&#10;\theta =\dfrac{mg{{R}^{2}}}{l}\sin \left( \dfrac{l}{R} \right)$&#10;&#10;Hence,&#10;potential energy is $\dfrac { m g R ^ { 2 } } { l } \sin \left( \dfrac {&#10;l } { R } \right)$&#10;&#10;

Question 8

Water stored in a dam possesses:

Accepted Answer

The potential energy&#10;possessed by the object is the energy present&#10;in it by virtue of its position or configuration. Water stored in a dam , when allowed to flow, has kinetic energy which was earlier stored as potential energy.

Question 9

Define potential energy. Derive an expression for gravitational potential energy. Mention its unit.

Accepted Answer

The energy possessed by a body due to its position or configuration is called potential energy.Consider a body of mass $&#8216;m&#8217;$ lying on the surface of the earth. Force required to lift the body is equal to its weight$\therefore F = mg$Let it be lifted through a height $&#8216;h&#8217;$,The work done on the body is $W = FS$$W = mgh (\because S = h)$This work done is stored in the body as its potential energy,$\therefore E_{p} = mgh$i.e. Gravitational potential energy $= mgh$Its unit is Joule.

Question 10

A body of mass 0.2 kg falls from a height of 10m to a height of 6m above the ground. Find the loss in potential energy taking place in the body. [$g=10ms^{-2}$].

Accepted Answer

Given : $m= 0.2 kg, g= 10 ms^{-2}, h_1=10m, h^2=6m$.Loss in P.E. $=mg(h_1-h_2)$$= 0.2 \times 10 \times 10(10-6)$$=0.2 \times 10 \times 4=8 J$

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