Question

The kinetic energy of a particle increases continuously with time . Then select correct alternative. ?

(A) the magnitude of its linear momentum is increasing continuously.
(B) its height above the ground must continuously decrease.
(C) the work done by all forces acting on the particle must be positive.
(D) the resultant force on the particle must be parallel to the velocity at all times.

PLEASE CLEAR ALL POINS ( Specially D)

Answer 1

Given,Kinetic energy(K.E) of a particle increases with time(`t`),

So,we can write, `1/2 mv^2 =kt` (as Kinetic energy = `1/2 mv^2` and here `k` is a constant)

So,`v=sqrt((2kt)/m)` or, `(dx)/(dt) = sqrt((2k)/m) sqrt(t)`

or, `dx = sqrt((2k)/m) sqrt(t) dt`

Integrating we get, `x=2/3sqrt((2k)/m)( t)^(3/2)`

So,distance covered by the particle increases with time...b is not correct

Now, if momentum of the parcticle is `P` then kinetic energy is written as `P^2/(2m)`

So,we can write, `P^2/(2m) = kt`

So, `P prop sqrt(t)`

So, option a is correct.

Now, `P= sqrt(2mkt)`

So,force acting i.e `(dP)/(dt) =-2sqrt(2mk) *(1/sqrt(t))`

But, `x=2/3sqrt((2k)/m)( t)^(3/2)`

So, `F.x` i.e work done is not always positive(see the negative sign infront of the force expression)....C not correct.

Again,in its pathway, The force needs not to act parallel to velocity,it can act at an angle more than `0` degrees but less than `90` degrees, as in such case,the component of force in the direction of velocity will be enough to increase its kinetic energy.

But,if the force acts perpendicularly to the direction of velocity,there will be no change in kinetic energy i.e it will become constant and it can't be more than `90` degrees as well because see the expression of force,it has a negative sign infront of it,and decreases with time,so ultimately,force increases with time,but if the angle will be more than `90` degrees,cos theta will ne negative,and that will make the value positive once again,and then the force will decrease with increase in time......so option d is incorrect as well.