A 2.0 kglunchbox is sent sliding over a frictionless surface, in the positive direction of an xaxis along the surface. Beginning at time t=0, a steady wind pushes on the lunchbox in the negative direction of the xaxis. Figure 7-51shows the position xof the lunchbox as a function of time tas the wind pushes on the lunchbox. From the graph, estimate the kinetic energy of the lunchbox at (a) t=1.0 sand (b)t=5.0 s. (c) How much work does the force from the wind do on the lunchbox fromt=1.0 s to t=5.0 s?

It is given that,

Mass of the lunchbox is$m=2.0\mathrm{kg}$

Figure 7-51 is the graph of x Vst

The problem deals with the work done which is the fundamental concept of physics.Work is the displacement of an object when force is applied on it. Also, it involves the kinetic energy of an object which possesses due to motion. Usingthekinematic equation, the initial velocity and acceleration can be found. Further, using initial velocity and acceleration, velocities at given times can be calculated. Also, find the kinetic energies at that given points. Work done by the wind isthechange in kinetic energies of the box.

Formulae:

Kinetic energy is given by,

$\mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2$

Work done is given by,

$W=Fd$

Displacement in kinematic equation is given by,

$x=v_{1}t+\frac{1}{2}a{t}^2$

Where,

F is force, x, dare displacements, $v_i$is the initial velocitie, t is time, m is the mass, a is an acceleration.

Write the position as a function of time,

$x\left(t\right)=v_1\left(t\right)t+\frac{1}{2}a\left(t\right)t^2$

Let’s take two points on the graph to find the acceleration and the initial velocity of the lunchbox.

At t = 1.5s, the position is 1.25m, and at t = 8s,the position is 1.5m.

Using these values, write two equations.

For first position, write,

$$\begin{gathered} 1.25=v_1\left(1.5\right)+\frac{1}{2}a{\left(1.5\right)}^2 \\ 1.5=v_1\left(8\right)+\frac{1}{2}a{\left(8\right)}^2 \end{gathered}$$

Now, solve the above equations simultaneously for ‘a’,

a=-0.198

=-0.2${\mathrm{m}/\mathrm{s}}^2$

Plug in this value in one of the above equations,

$$\begin{gathered} 1.25=v_i(1.5)+\frac{1}{2}(-0.2){(1.5)}^2{v}_i=0.98 \\ =1.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, find velocity at t = 1.0s,

$$\begin{gathered} v_i=v_i+at \\ {v}_{f1}=1.0+(-0.2)(1.0) \\ =0.8\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the kinetic energy of the box at t = 1.0s is,

$$\begin{gathered} \mathrm{K}.\mathrm{E}=\frac{1}{2}\left(2.0\right){\left(0.8\right)}^2 \\ =0.64\mathrm{J} \\ \mathrm{K}.{\mathrm{E}}_1=0.6\mathrm{J} \end{gathered}$$\

Hence, kinetic energy of the lunchbox att= 1.0s is 0.6J.

Similarly, find velocity at t = 5.0s,

$$\begin{gathered} V_{f2}=1.0+\left(-0.2\right)\left(5.0\right) \\ {V}_{f2}=0\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the kinetic energy at t = 5.0s is $\mathrm{K}.\mathrm{E}=0\mathrm{J}$

Hence, kinetic energy of the lunchbox at t= 5.0s is 0J.

Work done by the wind is the change in kinetic energy of the box,

$$\begin{gathered} W=K.E_2-K.E_1 \\ =0-0.6 \\ W=-0.6\mathrm{J} \end{gathered}$$

Hence, work done by the wind on the lunchbox fromt = 1.0s to t = 5.0s is -0.6J.