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

Answer:

(a). The kinetic energy is 5.33 J

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

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

$\Delta k=W_{f}$

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

$K(x)=4x-\dfrac{x^3}{3}$....(I)

Now put the value of x in equation

$K(x)=4\times2-\dfrac{2^3}{3}$

$K(x)=5.33\ J$

The kinetic energy is 5.33 J

For K to be minimum

$\dfrac{dK}{dx}=0$

$4-\dfrac{3x^2}{3}=0$

$4-x^2=0$

$x=2$

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

For K to be maximum

$\dfrac{d^2K}{dx^2}=0$

$\dfrac{d^2K}{dx^2}=0-2x$

Put the value of x

$\dfrac{d^2K}{dx^2}=-4$

Now put the value of x in equation (I)

$K(x)=4\times2-\dfrac{2^3}{3}$

$K(x)=5.33\ J$

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.