Home
Class 12
PHYSICS
The potential energy of a 1kg particle f...

The potential energy of a 1kg particle free to move along x-axis is given by `V(x)=((x^(4))/(4)-(x^(2))/(2))J`. The total mechanical energy of the particle is 2J. Then, the maximum speed (in m/s) is :

A

`3//sqrt(2)`

B

`sqrt(2)`

C

`1//sqrt(2)`

D

2

Text Solution

Verified by Experts

The correct Answer is:
D
At mean position x = 0, PE. U = 0
`therefore` At `x=0`, K.E. = Total energy
`(1)/(2)mv^(2)=2impliesv=sqrt((2xx2)/(1))=2ms^(-1)`
Promotional Banner

Topper's Solved these Questions

  • ROTATIONAL MOTION

    MODERN PUBLICATION|Exercise LEVEL-II (MCQ)|74 Videos

  • ROTATIONAL MOTION

    MODERN PUBLICATION|Exercise LEVEL-III (MCQ)|10 Videos

  • RAY OPTICS

    MODERN PUBLICATION|Exercise RECENT COMPETITIVE QUESTION|32 Videos

  • SOLIDS & SEMICONDUCTOR DEVICES

    MODERN PUBLICATION|Exercise Revision Test|28 Videos

Similar Questions

Explore conceptually related problems

The speed nu of the particle moving along straight line is given by a+bv^(2)=x^(2) , where x is its distance from the origin. The acceleration of the particle is

The potential energy of a particle varies with distance x as U=(Ax^(1//2))/(x^(2)+B) where a and B are constants. The dimensional formula for AXB is :

The potential energy function for a particle executing linear simple harmonic motion is given by V(x)= kx^(2)"/"2 , where k is the force constant of the oscillator. For k= 0.5 Nm^(1) , the graph of V(x) versus x is shown in Fig. 6. 12. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches x= pm 2m .

The particle moves along the curve y=x^(2)+2x Then the point on the curve such that x and y coordiantes of the particle change with the same rate

The potential energy U of a body of mass m is given by U = ax + by where x and y are the position coordinates of the particle, the net force acting on the particle is: