A bead with mass $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{}\mathbf{kg}$ is moving along a wire in the positive direction of an x axis. Beginning at time t=0, when the bead passes through x=0 with speed 12 m/s, a constant force acts on the bead. Figure 7-24 indicates the bead’s position at these four times: t₀=0, t₁=1.0 s, t₂=2.0 s, and t₃=3.0 s. The bead momentarily stops at t=3.0 s. What is the kinetic energy of the bead at t=10 s ?

Using the formula ofthesecond kinematic equation, we can writetheequation of displacement forthegiven time and velocity. Then, by differentiating the displacement with respect to time, we can findthevelocity of the bead. Then we can find the kinetic energy of the bead using this velocity and mass.

Formula:

$$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_0+{\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{v}=\frac{d\mathrm{x}}{d\mathrm{t}} \\ \mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2 \end{gathered}$$

The position of a particle can be calculated as,

$\mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_0+{\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$

Here x₀ is the initial position, v₀ is the initial velocity, and a is the acceleration of the particle.

Substitute the values in the above equation, and we get,

$\mathrm{x}\left(\mathrm{t}\right)=\left(12\frac{\mathrm{m}}{\mathrm{s}}\right)\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$ (1)

When x=10 m, t-1.0 s the above equation will be,

$\begin{array}{rcl}10\mathrm{m}& =& \left(12\frac{\mathrm{m}}{\mathrm{s}}\right)\left(1.0\mathrm{s}\right)+\frac{1}{2}\mathrm{a}{\left(1.0\mathrm{s}\right)}^2\\ \mathrm{a}& =& -4.0\mathrm{m}/{\mathrm{s}}^2\end{array}$

Therefore, equation (1) can be written as,

$\mathrm{x}\left(\mathrm{t}\right)=\left(12\mathrm{m}/\mathrm{s}\right)\mathrm{t}-\left(2\mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2$

We know that velocity can be written in the form of position as,

$\mathrm{v}\left(\mathrm{t}\right)=\frac{d\mathrm{x}}{d\mathrm{t}}$

Substitute the values in the above equation, and we get,

$$\begin{gathered} \mathrm{v}\left(\mathrm{t}\right)=\frac{d\left[\left(12\mathrm{m}/\mathrm{s}\right)\mathrm{t}-\left(2\mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2\right]}{d\mathrm{t}} \\ \mathrm{v}\left(\mathrm{t}\right)=\left(12\mathrm{m}/\mathrm{s}\right)-\left(4\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t} \end{gathered}$$

At t=3.0 s, ${\mathrm{v}}_{\left(\mathrm{t}=3.0\mathrm{s}\right)}=0$,because the bead stops momentarily.

At t=10 s, velocity can be calculated as,

data-custom-editor="chemistry" $\begin{array}{rcl}\mathrm{v}\left(10.0\mathrm{s}\right)& =& 12\frac{\mathrm{m}}{\mathrm{s}}-4.0\frac{\mathrm{m}}{{\mathrm{s}}^2}\times \left(10.0\mathrm{s}\right)\\ & =& -28\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Therefore, kinetic energy can be calculated as,

$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$

Substitute the values in the above equation, and we get,

$$\begin{gathered} \mathrm{KE}=\frac{1}{2}\left(1.8\times {10}^{-2}\mathrm{kg}\right){\left(-28\frac{\mathrm{m}}{\mathrm{s}}\right)}^2 \\ \mathrm{KE}=7.1\mathrm{J} \end{gathered}$$

Therefore, the kinetic energy of the bead at t=10 s is 7.1 J.