Home
Class 12
PHYSICS
The potential energy of a particle of ma...

The potential energy of a particle of mass 5 kg moving in the `x-y` plane is given by `U=(-7x+24y)J`, where x and y are given in metre. If the particle starts from rest, from the origin, then the speed of the particle at `t=2`s is

Text Solution

Verified by Experts

The correct Answer is:
10
`vec-(deltaU)/(deltax)hati-(deltaU)/(deltay)hatj=hati-24 hatj`
`:. a_(x)=(F_(x))/(m)=(7)/(5)=1.4 ms^(-1)` along positive x-axis
`a_(y)=(F_(y))/(m)=-(24)/(5)`
`=4.8 ms^(-2)` along negative y-axis
`:. v_(x)=a_(x)t=1.4xx2`
`=-2.8 ms^(-2)`
`v_(uy)=4.8xx2=9.6 ms^(-1)`
`:. v=sqrt(v_(x)^(2)+v_(y)^(2))=10 ms^(-1)`
Promotional Banner

Topper's Solved these Questions

  • NTA TPC JEE MAIN TEST 100

    NTA MOCK TESTS|Exercise PHYSICS (SUBJECTIVE NUMERICAL)|10 Videos

  • NTA NEET TEST 98

    NTA MOCK TESTS|Exercise PHYSICS|45 Videos

  • NTA TPC JEE MAIN TEST 101

    NTA MOCK TESTS|Exercise PHYSICS |30 Videos

Similar Questions

Explore conceptually related problems

The potential energy of a particle of mass 1 kg moving in X-Y plane is given by U=(12x+5y) joules, where x an y are in meters. If the particle is initially at rest at origin, then select incorrect alternative :-

The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(x-4)J , where x is in meter. It can be concluded that

The potential energy of a partical of mass 2 kg moving in y-z plane is given by U=(-3y+4z)J where y and z are in metre. The magnitude of force ( in newton ) on the particle is :

The potential energy of a particle of mass 5 kg moving in the x-y plane is given by U=-7x+24y joule, x and y being in metre. Initially at t = 0 the particle is at the origin. (0, 0) moving with a velocity of 6[2.4hat(i)+0.7hat(j)]m//s . The magnitude of force on the particle is :

The potential energy of a particle of mass 5 kg moving in xy-plane is given as U=(7x + 24y) joule, x and y being in metre. Initially at t=0 , the particle is at the origin (0,0) moving with velovity of (8.6hati+23.2hatj) ms^(1) , Then

The potential energy of a particle of mass 1 kg in a conservative field is given as U=(3x^(2)y^(2)+6x) J, where x and y are measured in meter. Initially particle is at (1,1) & at rest then:

The potential energy of particle of mass 1kg moving along the x-axis is given by U(x) = 16(x^(2) - 2x) J, where x is in meter. Its speed at x=1 m is 2 m//s . Then,