Q49E - University Physics with Modern Physics

Question

A force in the $\mathbf{+}\mathit{x}$  direction with magnitude $\mathit{F}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{N}}\mathbf{-}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{530}\mathbf{ }\mathbf{\text{N/m)}}\mathit{x}$ is applied to a $\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{kg}}$ box that is sitting on the horizontal, frictionless surface of a frozen lake. $\mathit{F}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ is the only horizontal force on the box. If the box is initially at rest at $\mathit{x}\mathbf{=}\mathbf{0}$ , what is its speed after it has traveled $\mathbf{14}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{m}}$ ?

Step-by-Step Solution

Answer

$8.17 \text{m/s}$

Step 1: Identification of given data

Mass of block is $m=6.00 \mathrm{kg}$

Distance travelled by block is $14.0 \text{m}$

Step 2: Significance of law of conservation of energy

Energy cannot be created or destroyed; it can only be transformed from one form into another or transferred from one object to another, according to the physical rule of conservation of energy.

Step 3: Determining the speed

Firstly determine the work done and it is given as

$W={\int }_{x=0}^{x=14}F\left(x\right)dx$

Substitute all the value in above equation to get the value of work done

$\begin{array}{rcl}W& =& {\int }_{x=0}^{x=14}\left(18.0 \text{N}-(0.530 \text{N/m)}x\right)dx\\ & =& {\left[\left(18.0 \text{N}\right) x-(0.530 \text{N/m)}\frac{x^2}{2}\right]}_{x=0}^{x=14}\\ & =& 200.06 \text{J}\end{array}$

This work done will be equal to the change in kinetic energy by the law of energy conservation, so

$W=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$                                                                                                              ...(i)

Where $u$ is initial velocity and is the final velocity

It is given in the question that the initial velocity is zero so the initial velocity$u = 0$

Substitute all the values in equation (i)

$\begin{array}{rcl}200.06 \text{J}& =& \frac{1}{2} \times 6 \text{kg}\times v^2\\ v& =& 8.17 \text{m/s}\\ & & \end{array}$

Hence the speed is $8.17 \text{m/s}$