Question

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 $\overrightarrow{\mathbf{F}}\left(x\right)\mathbf{=}\left(2.5-x^2\right)\hat{\mathbf{i}}\mathbf{N}$ , 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?

Short answer

1. The kinetic energy of the block at $\mathrm{x}=2.0\mathrm{m},{\mathrm{K}}_{\mathrm{f}}=2.3\mathrm{J}$
2. The maximum kinetic energy of the block between x=0 and x=2.0 m is ${\mathrm{K}}_{\max }=2.6\mathrm{J}$

Step-by-step solution

Given

1. The mass of block is, $\mathrm{m}=1.5\mathrm{kg}$
2. The applied force, $\mathrm{F}=\left(2.5-{\mathrm{x}}^2\right)\hat{\mathrm{i}}\mathrm{N}$
3. The initial position of the block is x=0

Understand the concept

We can use the equation ofthework-energy theorem to find the kinetic energy at . To find the maximum value of kinetic energy, we first find the position at whichtheblock has maximum kinetic energy. Next, we can find its position by takingthederivative oftheequation of kinetic energy and setting it equal to zero. Hence, kinetic energy at that position will be maximum kinetic energy.

Formula:

The equation of work-energy theorem

$\mathbf{W}\mathbf{=}{\mathbf{K}}_{\mathbf{f}}\mathbf{-}{\mathbf{K}}_{\mathbf{i}}$

Condition for extrema,

$$\begin{gathered} \frac{\mathbf{d}\mathbf{K}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{0} \\ \mathbf{W}\mathbf{=}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{i}}}^{{\mathbf{x}}_{\mathbf{f}}}\mathbf{F}\mathbf{d}\mathbf{x} \end{gathered}$$

(a) Calculate the kinetic energy of the block as it passes through x=2.0 m

The equation of work-energy theorem is,

$\mathrm{W}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}$

As, the particle is at rest initially, ${\mathrm{K}}_{\mathrm{i}}=0$. So,

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{f}}& =& \mathrm{W}\\ & =& {\int }_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{F}d\mathrm{x}\end{array}$

Substitute the value of F in the above equation.

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{f}}& =& {\int }_0^{2.0}\left(2.5-{\mathrm{x}}^2\right)d\mathrm{x}\\ & =& {\left[2.5\mathrm{x}-\frac{{\mathrm{x}}^3}{3}\right]}_0^2\\ & =& 2.3\mathrm{J}\end{array}$

Therefore, kinetic energy of the block as it passes through x=2.0 m is 2.3 J

(b) Calculate the maximum kinetic energy of the block between x=0 and x=2.0 m

The condition for extreme is,

$\frac{d\mathrm{K}}{d\mathrm{x}}=0$

Solving this equation for$\mathrm{x}={\mathrm{x}}_{\mathrm{c}}$, we get

$2.5-{\mathrm{x}}_{\mathrm{c}}^2=0$

So,

${\mathrm{x}}_{\mathrm{c}}=1.6\mathrm{m}$

It means, the block will have maximum kinetic energy as$\mathrm{x}=1.6\mathrm{m}$

So,

$\begin{array}{rcl}{\mathrm{K}}_{\max }& =& {\int }_0^{1.6}\left(2.5-{\mathrm{x}}^2\right)d\mathrm{x}\\ & =& {\left[2.5\mathrm{x}-\frac{{\mathrm{x}}^3}{3}\right]}_0^{1.6}\\ {\mathrm{K}}_{\max }& =& 2.6\mathrm{J}\end{array}$

Therefore, the maximum kinetic energy of the block between x=0 and x=2.0 m is 2.6 J.