Question

g A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force along an x axis is applied to the block. The force is given by , 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

(a). The kinetic energy is 5.33 J

(b). The maximum kinetic energy is 5.33 J.

Step-by-step explanation:

Given that,

Mass of block = 1.5 kg

Suppose the force is

`F=(4-x^2)i\ N`

(a). We need to calculate the kinetic energy of the block

Using work energy theorem

`\Delta k=W_(f)`

`K(x)=\int_(0)^(x){4-x^2}dx`

`K(x)=4x-(x^3)/(3)`....(I)

Now put the value of x in equation

`K(x)=4*2-(2^3)/(3)`

`K(x)=5.33\ J`

The kinetic energy is 5.33 J

For K to be minimum

`(dK)/(dx)=0`

`4-(3x^2)/(3)=0`

`4-x^2=0`

`x=2`

(b). We need to calculate the maximum kinetic energy of the block

For K to be maximum

`(d^2K)/(dx^2)=0`

`(d^2K)/(dx^2)=0-2x`

Put the value of x

`(d^2K)/(dx^2)=-4`

Now put the value of x in equation (I)

`K(x)=4*2-(2^3)/(3)`

`K(x)=5.33\ J`

The maximum kinetic energy is 5.33 J

Hence, (a). The kinetic energy is 5.33 J

(b). The maximum kinetic energy is 5.33 J.