Q: What is the work to be done to increase the velocity of a car from \$30\ km\ h^{-1}\$ to \$60\ km\ h^{-1}\$ If the mass of the car is \$1500\ kg\$?

Initial velocity \$u = 30 \ km/h = 30\times \dfrac{5}{18} = \dfrac{25}{3} \ m/s\$Final velocity \$v = 60 \ km/h = 60\times \dfrac{5}{18} = \dfrac{50}{3} \ m/s\$Mass \$m = 1500 \ kg\$Work done \$W = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2\$\$\implies \ W = \dfrac{m}{2}(v^2-u^2) = \dfrac{1500}{2}\times ((\dfrac{50}{3})^2 - (\dfrac{25}{3})^2) = 156187 \ J\$

Q: Prove work energy theorem for a constant force.

Consider a force \$F\$ acting on a body of mass \$m\$. The body moves with an acceleration a work done in moving the body to distance, \$x=F.x=max\quad \longrightarrow \left( 1 \right) \$change in \$K.E\$ of body \$=\dfrac { 1 }{ 2 } { mV }^{ 2 }\$\${ V }^{ 2 }={ \mu }^{ 2 }+2as\Rightarrow { V }^{ 2 }=2ax\$\$\Rightarrow \Delta K.E=\dfrac { 1 }{ 2 } m.2ax= max \$....... same as \$(1)\$Thus Work done \$=\$ Change in Kinetic energy (Work energy theorem).

Q: For the path PQR in a conservative force field, the amount of work done in carrying a body from P to Q & from Q to R are \$5\$J & \$2\$J respectively. The work done in carrying the body from P to R will be?

Workdone in closed path is zero.\${ W }_{ PtoQ }+{ W }_{ QtoR }+{ W }_{ RtoP }=0\$\$2+5+{ W }_{ RtoP }=0{ W }_{ RtoP }\$\${ W }_{ RtoP }=-7\$\${ W }_{ PtoR }=-{ W }_{ RtoP }=-\left( -7 \right) =7J\$

Q: A block of mass m is attached with a massless spring of force constant k. The block is placed over a rough inclined surface for which the coefficient of friction is \$\mu = 3/4\$. The minimum value of M required to move the block up the plane is (Neglect mass of string and pulley and friction in pulley)

Let mass \$M\$ decends \$x\$ meter to lift the load then\$kx=mg\sin\theta+\mu mg\cos\theta\$\$kx=6m+6m\$\$kx=12m\$Now,Applying work energy theorm,\$W_{spring}+W_{gravity}=\$change in K.E\$-\dfrac1{2}kx^2+Mgx=0\$\$kx=2Mg\$\$12m=2Mg\$\$M=m\times \dfrac{3}{5}\$\$M=\dfrac{3m}{5}\$

Question 1

What is the work to be done to increase the velocity of a car from $30\ km\ h^{-1}$ to $60\ km\ h^{-1}$ If the mass of the car is $1500\ kg$?

Accepted Answer

Initial velocity    $u = 30 \ km/h = 30\times \dfrac{5}{18} = \dfrac{25}{3} \ m/s$Final velocity    $v = 60 \ km/h = 60\times \dfrac{5}{18} = \dfrac{50}{3} \ m/s$Mass   $m = 1500 \ kg$Work done   $W = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2$$\implies \ W = \dfrac{m}{2}(v^2-u^2)  = \dfrac{1500}{2}\times ((\dfrac{50}{3})^2 - (\dfrac{25}{3})^2) = 156187 \ J$

Question 2

State and prove the work-energy theorem.ORShow that work done by a moving body is equal to the change in K.E. of the body.

Accepted Answer

The work energy theorem states that work done on a body is equal to the net change in its energy.(P.E or K.E)Proof:Consider a body of mass $&#8216;m&#8217;$ moving with an initial velocity $u$. Let a constant force $F$ acting on a body changes its velocity to $v$. Let $s$ be the distance traveled.From the equation, $v^{2} = u^{2} + 2as$,we get $v^{2} &#8211; u^{2} = 2as$$\frac{1}{2}(v^{2} &#8211; u^{2}) = as$Multiplying both sides by $m$, we have$\frac{1}{2}m(v^{2} &#8211; u^{2}) = mas$$\frac{1}{2}m(v^{2} &#8211; u^{2}) = F.s(\because F = ma)$or $\frac{1}{2}mv^{2} &#8211; \frac{1}{2}mu^{2} = W$

Question 3

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to

Accepted Answer

Hint:Using the supplied data, derive a particle acceleration relationship. In this equation, substitute the value in the form of a change in velocity. Substitute the value of time change in this equation as well. Integrate this equation between the velocity and displacement limits to get the relationship between the change in kinetic energy of the particle and its displacement.Step 1:Find the loss of kinetic energy for any displacement $x$ is proportional to.Let,Net work done by all the forces give the change in kinetic energy$\Rightarrow$$W=\frac{1}{2} m v^{2}-\frac{1}{2} m v_{0}^{2}$ $W=k_{f}-k_{i}$where,$m=\text { mass of the body }$$v_{0}=\text { initial velocity }$$v=\text { final velocity }$Step 2:Retardation $a \alpha-x$$\Rightarrow$$ \frac{d v}{d t}=-k x \quad[\mathrm{k}$ is proportionality constant$\Rightarrow$$ v \frac{d v}{d x}=-k x$$\Rightarrow$$\int_{v_{1}}^{v_{2}} v d v=-k \int_{0}^{x} x d x$$\Rightarrow \frac{1}{2}\left(v_{2}^{2}-v_{1}^{2}\right)=-\frac{1}{2} k x^{2}$Loss in kinetic energy $=\frac{1}{2} m\left(v_{2}^{2}-v_{1}^{2}\right)=-\frac{1}{2} k x^{2}$Loss in kinetic energy $\alpha  x^{2}$Therefore, The correct option is 'C' is $x^{2}$.

Question 4

State work-energy theorem. Prove it for a variable force.

Accepted Answer

Work energy theorem -It states that the work done by the net force acting on a body is equal to the changed produced in kinetic energy of the body. Let F be the variable force $\therefore$ Work done by this variable force , $W=\int_{x_i}^{x_f} F\cdot dx$where $x_i$ is the initial position and $x_f$ is final position.Also Kinetic energy of an object , $K=\dfrac 12 mv^2$ $\implies \dfrac{dK}{dt}=mv \dfrac{dv}{dt}$$\implies \dfrac{dK}{dt}= ma\dfrac{dx}{dt}$$\implies \dfrac{dK}{dt}F\dfrac{dx}{dt}$$\implies dK=Fdx$$\implies \int _{K_i}^{K_f}dK=\int_{x_i}^{x_f} F\cdot dx$$\implies \Delta K= W$Where $\Delta K $ is the change in kinetic energy

Question 5

State and derive work energy theorem.

Accepted Answer

Work energy theorem states that the change in kinetic energy of an object is equal to the net-work done on it by the net force.Let us suppose that a body is initially at rest and a force. $\vec F $is applied on the body to displace it through$ d \vec  s$along the direction of the force. Then, a small amount of work done is given bydw = $\vec F . d \vec s $ = FdsAlso, according to Newton's second law of motion, we haveF = ma where a is acceleration produced (in the direction of force) on applying the force. Therefore, dw = Mada = $M \dfrac{dv}{dt} ds$Now, work done by the force in order to increase its velocity from u (initial velocity) to v (final velocity) is given byW = $\int_u^v Mvdv $ = $M \int_u^v vdv$     = $M|\dfrac{v^2}{2}|_u^v$W = $\dfrac{1}{2} Mv^2 - \frac{1}{2}Mu^2$Hence, work done on a body by a force is equal to the change in its kinetic energy.

Question 6

Prove work energy theorem for a constant force.

Accepted Answer

Consider a force $F$ acting on a body of mass $m$. The body moves with an acceleration a work done in moving the body to distance, $x=F.x=max\quad \longrightarrow \left( 1 \right) $change in $K.E$ of body $=\dfrac { 1 }{ 2 } { mV }^{ 2 }$${ V }^{ 2 }={ \mu  }^{ 2 }+2as\Rightarrow { V }^{ 2 }=2ax$$\Rightarrow \Delta K.E=\dfrac { 1 }{ 2 } m.2ax= max $....... same as $(1)$Thus Work done $=$ Change in Kinetic energy (Work energy theorem).

Question 7

For the path PQR in a conservative force field, the amount of work done in carrying a body from P to Q & from Q to R are $5$J & $2$J respectively. The work done in carrying the body from P to R will be?

Accepted Answer

Workdone in closed path is zero.${ W }_{ PtoQ }+{ W }_{ QtoR }+{ W }_{ RtoP }=0$$2+5+{ W }_{ RtoP }=0{ W }_{ RtoP }$${ W }_{ RtoP }=-7$${ W }_{ PtoR }=-{ W }_{ RtoP }=-\left( -7 \right) =7J$

Question 8

A block of mass m is attached with a massless spring of force constant k. The block is placed over a rough inclined surface for which the coefficient of friction is $\mu = 3/4$. The minimum value of M required to move the block up the plane is (Neglect mass of string and pulley and friction in pulley)

Accepted Answer

Let mass $M$ decends $x$ meter to lift the load then$kx=mg\sin\theta+\mu mg\cos\theta$$kx=6m+6m$$kx=12m$Now,Applying work energy theorm,$W_{spring}+W_{gravity}=$change in K.E$-\dfrac1{2}kx^2+Mgx=0$$kx=2Mg$$12m=2Mg$$M=m\times \dfrac{3}{5}$$M=\dfrac{3m}{5}$

Question 9

What are conservative and non conservative forces ?

Accepted Answer

Conservative Forces :Conservative forces are those for which work done depends only on initial and final points.For example, Gravitational Force, Electrostatic ForceNon-Conservative Forces :Non-Conservative forces are those where the work done or the kinetic energy did depend on other factors such as the velocity or the particular path taken by the object.For example, Frictional Force

Question 10

An object of mass, m is moving with a constant velocity v. How much work should be done on the object in order to bring the object to rest ?

Accepted Answer

An object with mass 'm' moving with velocity 'V' has kinetic energy of  $\dfrac{1}{2}m{v}^{2}$ . In order to bring it to rest, its velocity has to be reduced to zero, and in order to accomplish that, the kinetic energy has to be drained off and sent somewhere else. An external force has to absorb energy from the object, i.e. do negative work on it, equal to its kinetic energy,or$-\dfrac{1}{2}m{v}^{2}$.

Work-Energy Theorem

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