91影视

The given data is listed below as-

• The mass of the watermelon is $m=4.80\mathrm{kg}$
• The initial velocity of watermelon is, $V_1=0\mathrm{m}/\mathrm{s}$
• The height of the roof is $s=18\mathrm{m}$

The kinetic energy of a particle equals the amount of work required to accelerate the particle from rest to speed V. Therefore, kinetic energy on the particle is given by-

$\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mV}}^{\mathbf{2}}$

The kinetic energy is a scalar and it is always positive or zero.

(a)

The work done by gravity on watermelon during its displacement from the roof to the ground is calculated by the following method:

$W_g=mgs$

Here, m is the mass of the watermelon, g is the gravitational constant, and s is the height of the roof.

For, $m=4.80\mathrm{kg},g=9.8{\mathrm{ms}}^{鈥�2}$and $s=18\mathrm{m}$

The work done will be

$$\begin{gathered} W_g=mgs \\ =4.8\mathrm{kg}脳9.8{\mathrm{ms}}^{鈥�2}脳18\mathrm{m} \\ =846.72\mathrm{J} \end{gathered}$$

Thus, the work done by gravity on the watermelon during its displacement from the roof to the ground is $846.72\mathrm{J}$

(b)(i)

From the work-energy theorem,

$W_{tot}=\frac{1}{2}m{V^2}_{2B}-\frac{1}{2}m{V^2}_{1A}$

The initial velocity of watermelon is 0. Therefore, obtain the new equation as below:

$$\begin{gathered} W_{tot}=\frac{1}{2}m{V^2}_{2B}-0 \\ {W}_{tot}=K_{2B} \end{gathered}$$

Therefore, $K_{2B}=846.72\mathrm{J}$

Thus, watermelon鈥檚 kinetic energy before watermelon strikes the ground is 846.72 J

(b)(ii)

The velocity of a particle at time t with constant acceleration is,

role="math" localid="1664857957331" $V^2={V^2}_0+2as$鈥�..(1)

Now, the object is falling with constant acceleration along y-axis and $S=y_2-y_1$

So, equation (1) will become

${V^2}_{2B}={V^2}_{1A}-2g\left(y_2-y_1\right)$

Here, initial velocity, $V_{1A}=0,g=9.8{\mathrm{ms}}^{鈥�2}$and $S=y_2-y_1=18\mathrm{m}$

Therefore,

$$\begin{gathered} {V^2}_{2B}=0-2脳9.8{\mathrm{ms}}^{鈥�2}脳18\mathrm{m} \\ {V^2}_{2B}=362.8{\mathrm{m}}^2{\mathrm{s}}^{鈥�2} \end{gathered}$$

Hence, $V_{2B}=362.8{\mathrm{ms}}^{鈥�1}$

Thus, speed of watermelon just before it strikes the ground is $362.8{\mathrm{ms}}^{鈥�1}$.

(c)

The free-body diagram for an extra force due to air resistance that will exert on watermelon due to gravity

(b) Now the total force of object under free fall due to existence of air resistance will be

$F_{total}=F_g-F_{air}$

Therefore, the kinetic energy and speed of watermelon will change accordingly due to existence of air resistance. The new kinetic energy will be less than that before there was no air resistance.

Thus, Part (a) will remain unchanged and both parts (i) and (ii) of part (b) will be different if there were appreciable air resistance.