Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Work Power and Energy question

2017 · Shift 0 · Q65
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Work Power and Energy
  5. /2017 · Shift 0 · Q65

Work Power and Energy question

2017 · Shift 0 · Q65

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is 18mv02{1 \over 8}mv_0^281​mv02​, the value of k will be:
  1. A
    10-1 kg m-1 s-1
  2. B
    10-3 kg m-1
  3. C
    10-3 kg s-1
  4. D
    10-4 kg m-1
View written solutionFree

Correct answer: D

  1. Given data
  • Mass: m=10−2 kgm = 10^{-2}\,\text{kg}m=10−2kg
  • Retarding force: F=−kv2F = -k v^2F=−kv2
  • Initial speed: v0=10 m s−1v_0 = 10\,\text{m s}^{-1}v0​=10m s−1
  • After t=10 st=10\,\text{s}t=10s, energy becomes 18mv02\dfrac{1}{8} m v_0^281​mv02​
  1. Find the speed after 10 s from energy condition

Initial kinetic energy is K0=12mv02K_0 = \frac12 m v_0^2K0​=21​mv02​

After 10 s, kinetic energy is given as K=18mv02K = \frac18 m v_0^2K=81​mv02​

But also, K=12mv2K = \frac12 m v^2K=21​mv2

So, 12mv2=18mv02\frac12 m v^2 = \frac18 m v_0^221​mv2=81​mv02​

Cancelling mmm, 12v2=18v02\frac12 v^2 = \frac18 v_0^221​v2=81​v02​ v2=14v02v^2 = \frac14 v_0^2v2=41​v02​ v=v02=5 m s−1v = \frac{v_0}{2} = 5\,\text{m s}^{-1}v=2v0​​=5m s−1

  1. Set up equation of motion

Since the only force is resistive, mdvdt=−kv2m\frac{dv}{dt} = -k v^2mdtdv​=−kv2

Rearrange: dvv2=−kmdt\frac{dv}{v^2} = -\frac{k}{m} dtv2dv​=−mk​dt

  1. Integrate using limits

Initially, at t=0t=0t=0, v=v0=10v=v_0=10v=v0​=10. After t=10 st=10\,\text{s}t=10s, v=5v=5v=5.

So, ∫105dvv2=−km∫010dt\int_{10}^{5} \frac{dv}{v^2} = -\frac{k}{m} \int_0^{10} dt∫105​v2dv​=−mk​∫010​dt

Now, ∫dvv2=∫v−2dv=−1v\int \frac{dv}{v^2} = \int v^{-2} dv = -\frac{1}{v}∫v2dv​=∫v−2dv=−v1​

Therefore, [−1v]105=−km(10)\left[-\frac{1}{v}\right]_{10}^{5} = -\frac{k}{m}(10)[−v1​]105​=−mk​(10)

−15−(−110)=−10km-\frac{1}{5} - \left(-\frac{1}{10}\right) = -\frac{10k}{m}−51​−(−101​)=−m10k​

−15+110=−10km-\frac{1}{5} + \frac{1}{10} = -\frac{10k}{m}−51​+101​=−m10k​ −110=−10km-\frac{1}{10} = -\frac{10k}{m}−101​=−m10k​

110=10km\frac{1}{10} = \frac{10k}{m}101​=m10k​ k=m100k = \frac{m}{100}k=100m​

Substitute m=10−2 kgm=10^{-2}\,\text{kg}m=10−2kg: k=10−2100=10−4k = \frac{10^{-2}}{100} = 10^{-4}k=10010−2​=10−4

  1. Unit of kkk

From F=kv2F = kv^2F=kv2, [k]=[F][v2]=kg m s−2m2s−2=kg m−1[k] = \frac{[F]}{[v^2]} = \frac{\text{kg m s}^{-2}}{\text{m}^2\text{s}^{-2}} = \text{kg m}^{-1}[k]=[v2][F]​=m2s−2kg m s−2​=kg m−1

So, k=10−4 kg m−1k = 10^{-4}\,\text{kg m}^{-1}k=10−4kg m−1

  1. Check options
  • A: 10−1 kg m−1s−110^{-1}\,\text{kg m}^{-1}\text{s}^{-1}10−1kg m−1s−1 ❌ wrong unit/value
  • B: 10−3 kg m−110^{-3}\,\text{kg m}^{-1}10−3kg m−1 ❌ wrong value
  • C: 10−3 kg s−110^{-3}\,\text{kg s}^{-1}10−3kg s−1 ❌ wrong unit/value
  • D: 10−4 kg m−110^{-4}\,\text{kg m}^{-1}10−4kg m−1 ✅ correct

Final answer: Option D

PreviousNext

More from Work Power and Energy

  • A car of weight W is on an inclined road that rises by 100 m over a distance of 1 km and applies a constant frictional force 20W​ on the car. While moving uphill on the road at a speed of 10 ms−1, the car needs power P. If it…2016 · MCQ
  • Velocity-time graph for a body of mass 10 kg is shown in figure. Work-done on the body in first two seconds of the motion is : Includes diagram2016 · MCQ
  • A particle of mass M is moving in a circle of fixed radius R in such a way that its centripetal acceleration at time t is given by n2 R t2 where n is a constant. The power delivered to the particle by the force acting on it, is :2016 · MCQ
  • A point particle of mass m, moves long the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals μ. The particle is released, from rest from the point P… Includes diagram2016 · MCQ
  • A person trying to lose weight by burning fat lifts a mass of 10kg upto a height of 1m1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done…2016 · MCQ
  • When a rubber-band is stretched by a distance x, it exerts restoring force of magnitude F=ax+bx2 where a and b are constants. The work done in stretching the unstretched rubber-band by L is :2014 · MCQ
  • This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements. If two springs S1​ and S2​ of force constants k1​ and k2​,…2012 · MCQ
  • The potential energy function for the force between two atoms in a diatomic molecule is approximately given by U(x)=x12a​−x6b​, where a and b are constants and x is the distance…2010 · MCQ