Home
Class 12
PHYSICS
A particle of mass m is subjected to a f...

A particle of mass m is subjected to a force `vecF=F_(0)[cos (t)hati+sin(1)hatj]`. If initially (t=0) the particle was at rest, the kinetic energy of the particle as a function of time is given by:

A

`(F_(0)^(2))/(m)(1-cos(2t))`

B

`(F_(0)^(2))/(m)(1-cost)`

C

`(F_(0)^(2))/(m) sin(t)`

D

`(F_(0)^(2))/(m)t`

Text Solution

Verified by Experts

`veca=(vecF)/(m)=(F_(0))/(m) [cos(t)hati+sin(t)hatj]`
`vecv=int_(0)^(t)vecadt=(F_(0))/(m)[sin(t)hati+(1-cos t)hatj]`
`KE=(1)/(2)m(vecv.vecv)=(1)/(2)m[(F_(0)^(2))/(m^(2))(2-2cos t)]`
`=(F_(0)^(2))/(m) (1-cos t)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • TEST PAPERS

    BANSAL|Exercise PHYSICS PART (A)|32 Videos

  • TEST PAPERS

    BANSAL|Exercise PHYSICS PART (B)|3 Videos

  • TEST PAPERS

    BANSAL|Exercise PHYSICS PART- C|4 Videos

  • SEMICONDUCTORS

    BANSAL|Exercise CBSE Question|32 Videos

Similar Questions

Explore conceptually related problems

A particle of unknown mass is acted upon by a force vecF=(100e^(-2t)hati) N . If at t = 0.00 s the particle is at rest, for the time interval t=0.00 s to t= 2.00 s find (a) the impulse on the particle and (b) the average force on the particle.

A particle of mass m kept at the origin is subjected to a froce vecF=(pt-qx)hati where t is the time elapsed and x is the x co-ordinate of the position of the particle. Particle starts its motion at t=0 with zero initial velocity. If p and q are positive constants, then:

Knowledge Check

  • A particle of mass m is released from rest and follows a particle as a function of time?

    A
    B
    C
    D

  • At t = 0 aparticle of mass m start moving from rest due to a force vecF = F_(0) sin (omega t)hati :

    A
    Particle perform `SHM` about its initial position of rest
    B
    Particle perform `SHM` with initial position as extreme position with angular frequency `omega`
    C
    At any instant, distance moved by the particle equals its displacement, from the initial position
    D
    initial velocity of particle increases with time but after time `t = 2 pi// omega` it becomes constant

  • A particle of mass m is acted upon by a force F= t^2-kx . Initially , the particle is at rest at the origin. Then

    A
    its displacement will be in simple harmonic
    B
    its velocity will be in simple harmonic
    C
    its acceleration will be in simple harmonic
    D
    None of the above

    • Similar Questions

    Explore conceptually related problems

    A particle of mass m at rest is acted upon by a force F time t. Its kinetic energy after the interval t is

    A particle of mass m at rest is acted upon by a force F for a time t. Its Kinetic energy after an interval t is

    A particle of mass mat rest is acted upon by a force F for a time t. Its kinetic energy after an interval t is:

    For a particle moving on a straight lint the variation of acceleration with time is given by the graph as shown. Initially the particle was at rest. Then the corresponding kinetic energy of the particle versus time graph will be .

    A particle of mass m at rest is acted upon by a force P for a time t. Its kinetic energy after an interval t is

    Home
    Profile