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Question: A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/s. It is subjected to a...

A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/s. It is subjected to a retarding force F=0.1 x J/mF = 0.1\,x\,J/m during its travel form x=20 mx = 20\,m to x=30 cmx = 30\,cm . Its final kinetic energy will be:
(a) 475 J
(b) 450 J
(c) 275 J
(d) 250J

Explanation

Solution

Some of the work is done by the retarding force applied on block which is equal to change in kinetic energy according to work energy theorem. From this, final kinetic energy can be found.

Complete step by step answer:
According to work energy theorem amount of work done is equal to change in kinetic energy.
Work done=[final  kinetic  energy]−[initial  kinetic  energy]Work\,done = \left[ {final\,\,kinetic\,\,energy} \right] - \left[ {initial\,\,kinetic\,\,energy} \right] … (i)
Let initial kinetic energy=12mv2 = \dfrac{1}{2}m{v^2} … (ii)
∴  m(mass)=10 kg\therefore \,\,m\left( {mass} \right) = 10\,kg
v (initial  speed)=10m/sv\,\left( {initial\,\,speed} \right) = 10m/s
Use given value in equation (ii)
(K.E.)initial=12(10) (10)2{\left( {K.E.} \right)_{initial}} = \dfrac{1}{2}\left( {10} \right)\,{\left( {10} \right)^2}
(K.E.)initial=500 joule{\left( {K.E.} \right)_{initial}} = 500\,joule … (iii)
To calculate final kinetic energy need to calculate work done by using the calculate method.
Let a small amount of work be done displacing the particle by dx.
dw=Fdx  cos⁡ Qdw = Fdx\,\,\cos \,Q … (iv)
∴  F=0.1 x J/m\therefore \,\,F = 0.1\,x\,J/m
Q=180oQ = {180^o}(force is retarding)
Hencecos⁡180o=−1\cos {180^o} = - 1
dw=−0.1 x dxdw = - 0.1\,x\,dx. … (v)
Integrate equation (v) by using limits of displacementx1=20m{x_1} = 20m to x2=30m{x_2} = 30m
∫dw=−∫x1=20x2=300.1 x.dx.\int {dw} = - \int\limits_{{x_1} = 20}^{{x_2} = 30} {0.1\,x.dx.}
w=− 0.1∣x22∣x1=20x2=30w = - \,0.1\left| {\dfrac{{{x^2}}}{2}} \right|_{{x_1} = 20}^{{x_2} = 30}
=− .05[(30)2−(20)2]= - \,.05\left[ {{{\left( {30} \right)}^2} - {{\left( {20} \right)}^2}} \right]
=− .05[900−400]= - \,.05\left[ {900 - 400} \right]
=− .05[500]= - \,.05\left[ {500} \right]
W=− 25 JouleW = - \,25\,Joule … (vi)
Put the value's of initial kinetic energy and work done in equation (i)
−25=(K.E.)final−500- 25 = {\left( {K.E.} \right)_{final}} - 500
(K.E.)final=−25+500{(K.E.)_{final}} = - 25 + 500
(K.E.)final=475  joule.{(K.E.)_{final}} = 475\,\,joule.

Note: Under retarding force Kinetic energy decreases this is due to fact that the retarding force results in decrease in velocity. That is why even the work done by retarding force comes out to be negative.