A body of mass 3 kg is under a constant force which causes a displacement s in metres in it,given by the relation \$s = \dfrac{1} {2} t^{2}\$ , where t is in seconds. Work done by the force in 2 seconds is:-

By using work energy theoremW = \$\Delta\$KE = \$\dfrac{1} {2} mv_{2}^{2}\$ - \$\dfrac{1} {2} mv_{1}^{2}\$Now s = \$\dfrac{1} {3} t^{2} \Rightarrow\$ v = \$\dfrac{2} {3} t\$\$\Rightarrow\$ v\$_{1}\$ = 0, v\$_{2}\$ = \$\dfrac{2} {3} \times \$ 2 = \$\dfrac{4} {3}\$ ms\$^{-1}\$ Therefore W = \$\dfrac{1} {2} \times\$ 3\$\left (\dfrac{4} {3}\right )^{2}\$ - \$\dfrac{1} {2} \times\$ 3 \$\times \left (0\right )^{2}\$ = \$\dfrac{8} {3}\$

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>>Class 11
>>Physics
>>Work, Energy and Power
>>Work Done by a Variable Force
>> $Work Done by a Variable Force$

Work Done by a Variable Force

Q: The upper half of an inclined plane of inclination \$\theta\$ is perfectly smooth while the lower half rough. A block starting from rest at the top of the plane will again come to rest at the bottom if the coefficient of friction between the block and the lower half of the plane is given by.

Suppose length of plane is L. When block descends the plane, rise in kinetic energy \$=\$ fall in potential energy \$=mg L\sin \theta i\$Work done by friction \$=\mu \times (reaction)\times \$distance \$0+\mu(mg \cos\theta )\times (L/2)\$\$=\mu(mg\cos \theta)\times (L/2)\$Now, work done\$=\$ change in KE\$mgL\sin \theta =\mu (mg \cos\theta )\times (L/2)\$\$\Rightarrow \tan\theta =\mu/2\$\$\therefore \mu =2\tan\theta\$.

Q: A position dependent force \$ F = 7-2x + 3x^2 \$ newton (where x is in meter) on a small body of mass 2 Kg & displace it from \$ x = 0\$ to \$ x = 5 \$ m. Calculate the work done.

\$\textbf{Step 1: Work done}\$Work done for small displacement can be defined as: \$ dW=Fdx \$ \$....(1)\$\$\textbf{Step 2: Integrate the equation}\$Total work done can be calculated by integrating the equation \$(1)\$ with limits \$x=0\$ to \$x=5m\$ \$ \int{dW}=\int\limits_{0}^{5}{Fdx} \$ \$ \int{dW}=\int\limits_{0}^{5}{\left( 7-2x+3{{x}^{2}} \right)dx} \$ \$\Rightarrow W=\left[ 7x-\dfrac{2{{x}^{2}}}{2}+\dfrac{3{{x}^{3}}}{3} \right]_{0}^{5} \$ \$\Rightarrow W=35-25+125= 135\ J \$Hence, the work done is \$135\ J\$

Q: Derive the formula for kinetic energy.

Let an object of mass m, move with uniform velocity u. Let us displace it by s, due to constant force F, acting on it. Work done on the object of mass 'm' is \$W=F \times s\$ ...(i) Due to the force, velocity changes to v and the acceleration produced is 'a'. Relationship between v, u, a and s can be given by formula \$v^2- u^2 = 2as \$\$\therefore s=\dfrac{v^2-u^2}{2a}\$..........(ii)\$F=ma\$............(iii)Substituting (ii) and (iii) in (i) we get\$W =F \times s\$\$=ma\times \dfrac{v^2-u^2}{2a}\$\$W=\dfrac{1}{2}m(v^2-u^2)\$If u=0,(object starts at rest)\$W=\dfrac{1}{2}mv^2\$Work done=Change in kinetic energy\$\therefore E_k=\dfrac{1}{2}mv^2\$

Q: An object of Mass \$40\ kg\$ is raised to a height of \$5m\$ above the ground what is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half way down.

Given : Mass of the object \$M = 40 kg\$ Height upto which it is raised \$h = 5\$ mPotential energy \$P.E = Mgh\$\$P.E = 40 \times 10 \times 5 = 2000 J\$Let its kinetic energy be \$E_k\$ when it is half way down i.e \$h' = \dfrac{h}{2}\$According to conservation of energy, \$K.E_i + P.E_i = K.E_f + P.E_f\$\$0+ 2000 = E_k + Mg\dfrac{h}{2}\$\$\therefore\$ \$2000 = E_k + 40 \times 10 \times \dfrac{5}{2}\$\$\implies E_k = 1000 J\$

Q: Consider a drop of rain water having mass \$1 g\$ falling from a height of \$1km\$. It hits the ground with a speed of \$50m/s\$. Take '\$g\$' constant with a value \$10m/s^2\$. The work done by the (i) gravitational force the (ii) resistive force of airs is:

\$m=1g=\dfrac{1}{1000}kg\$ \$h=1km=1000m\$work done by gravitional force \$w_g-mgh\$\$wg = \dfrac{1}{1000}\times 10\times 1000\$Change in kinetic energy \$\triangle K.E = \dfrac{1}{2}mv^2-0\$\$=\dfrac{1}{2}\dfrac{1}{1000}\times 50 \times 50-0\$\$=1.25J\$Work energy theorem: \$w_g+w_{air\, resistance} =\triangle K.E\$\$10J + resistance = 1.25-10\$where resistance\$ = -8.75J\$

Q: \$300 J\$ of work is done in sliding a \$2 kg\$ block up an inclined plane of height \$10 m\$. The work done against friction is (Take \$g = 10m/s\$ \$^2\$)

\$\textbf{Step 1: Draw the free body diagram }\$ \$\textbf{Step 2: Write the equation for work done }\$Let \$W_f\$ be the work done against frictionThe total work done on a particle is equal to the change in its mechanical energy\$W =W_f + PE\$ \$=W_f + mgh........(1)\$\$\textbf{Step 3: Work done against the friction }\$From equation (1)\$W_f = W- mgh\$Substituting the values :\$W_f = 300 J -(2 kg\times10 m/s^2 \times 10m)= 100J\$Hence option \$(B)\$ is correct.

Q: A bullet fired into a fixed target loses half of its velocity after penetrating \$3\ cm\$. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion ?

Case \$I\$ :\$\displaystyle \left [ \frac{u}{2} \right ]^2-u^2=2.a.3\$or \$\displaystyle -\frac{3u^2}{4}=2.a.3\Rightarrow a=-\frac{u^2}{8}\$Case \$II\$ :\$\displaystyle 0- \left [ \frac{u}{2} \right ]^2=2.a.x\$ or \$\displaystyle -\frac{u^2}{4}=2\left [ -\frac{u^2}{8} \right ]\times x\$\$\Rightarrow \: x=1\:cm\$Alternative method : Let K be the initial energy and F be the resistive force. Then according to work-energy theorem,\$W=\Delta K\$i.e., \$\displaystyle 3F=\frac{1}{2}mv^2-\frac{1}{2}m\left [ \frac{v}{2} \right ]^2\$ \$\displaystyle 3F=\frac{1}{2}mv^2\left [ 1-\frac{1}{4} \right ]\$ \$\displaystyle 3F=\frac{3}{4}\left [ \frac{1}{2}mv^2 \right ]\$ ............(1)and \$\displaystyle Fx=\frac{1}{2}m\left [ \frac{v}{2} \right ]^2-\frac{1}{2}m(0)^2\$i.e., \$\displaystyle \frac{1}{4}\left [ \frac{1}{2}mv^2 \right ]=Fx\$ .........(2) Comparing eqns. (1) and (2) \$F=Fx\$or \$x=1\:cm\$

Q: A particle moves along the x- axis from x=0 to x=5 m under the influence of a force given by \$F=7-2x+3x^2\$.Work done in the process is

\$ \displaystyle w = \int F dx \$\$ \displaystyle =\int_{0}^{5}7-2x+3x^2 \,dx \$\$ \displaystyle = [7x-\frac{2x^2}{2}+\frac{3x^2}{3}]_{0}^{5}\$\$ \displaystyle = [7x-x^2+x^3]_{0}^{5}\$\$ \displaystyle = [7\times5-5^2+5^3]-[0-0+0]\$\$ \displaystyle = [35-25+125]-0\$\$ = 135J\$In this question, answer is 135 J not 70J

Q: A force acts on a \$3\$g particle in such a way that the position of the particle as a function of time is given by \$x= 3t -4t^2+t^3\$, where \$x\$ is in meters and \$t\$ is in seconds. The work done during the first \$4\$ second is :

We have,mass, \$m=3\ g=0.003\ kg\$\$x=3t-4t^2+t^3\$Now, \$v=\dfrac{dx}{dt}=3-8t+3t^2\Rightarrow dx=(3-8t+3t^2)dt\$\$\Rightarrow a=\dfrac{dv}{dt}=0-8+6t\$Now,\$dw=Fdx\$\$\Rightarrow dw=(ma)dx\$\$\Rightarrow dw=(0.003)(-8+6t)(3-8t+3t^2)dt\$\$\Rightarrow dw=(0.003)(18t^3-72t^2+82t-24)dt\$\$\Rightarrow w=(0.003)\displaystyle \int_{0}^{4}{(18t^3-72t^2+82t-24)dt}\$\$\Rightarrow w=0.003\times 176=0.528\ J\$\$\Rightarrow w=530\ mJ\$

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