Question

A 1.1 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F with arrow(x) = (2.4 − x2)i hat N, where x is in meters and the initial position of the block is x = 0.

(a) What is the kinetic energy of the block as it passes through x = 2.0 m?
(b) What is the maximum kinetic energy of the block between x = 0 and x = 2.0 m?

Answer 1

Answer with Explanation:

Mass of block=1.1 kg

Th force applied on block is given by

F(x)=
`(2.4-x^2)\hat{i}N`

Initial position of the block=x=0

Initial velocity of block=
`v_i=0`

a.We have to find the kinetic energy of the block when it passes through x=2.0 m.

Initial kinetic energy=
`K_i=(1)/(2)mv^2_i=(1)/(2)(1.1)(0)=0`

Work energy theorem:

`K_f-K_i=W`

Where
`K_f=`Final kinetic energy

`K_i`=Initial kinetic energy

`W=Total work done`

Substitute the values then we get

`K_f-0=\int_(0)^(2)F(x)dx`

Because work done=
`Force* displacement`

`K_f=\int_(0)^(2)(2.4-x^2)dx`

`K_f=[2.4x-(x^3)/(3)]^(2)_(0)`

`K_f=2.4(2)-(8)/(3)=2.13 J`

Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J

b.Kinetic energy =
`K=2.4x-(x^3)/(3)`

When the kinetic energy is maximum then
`(dK)/(dx)=0`

`(d(2.4x-(x^3)/(3)))/(dx)=0`

`2.4-x^2=0`

`x^2=2.4`

`x=\pm√(2.4)`

`(d^2K)/(dx^2)=-2x`

Substitute x=
`√(2.4)`

`(d^2K)/(dx^2)=-2√(2.4)<0`

Substitute x=
`-√(2.4)`

`(d^2K)/(dx^2)=2√(2.4)>0`

Hence, the kinetic energy is maximum at x=
`√(2.4)`

Again by work energy theorem , the maximum kinetic energy of the block between x=0 and x=2.0 m is given by

`K_f-0=\int_(0)^(√(2.4))(2.4-x^2)dx`

`k_f=[2.4x-(x^3)/(3)]^(√(2.4))_(0)`

`K_f=2.4(√(2.4))-((√(2.4))^3)/(3)=2.48 J`

Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J