Question

A 4.80-kg watermelon is dropped from rest from the roof of an 18.0-m tall building and feels no appreciable air resistance. (a) calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what is the watermelon’s (i) Kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be different if there were appreciable air resistance?

Short answer

(a)$W_g=846.72\mathrm{J}$

(b) (i) $K_{2B}=846.72\mathrm{J}$ and (ii) $V_{2B}=18.78{\mathrm{ms}}^{-1}$

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

Step-by-step solution

Identification of the given data

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 $s=18\mathrm{m}$

Significance of the kinetic energy

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.

Determination of the work done by gravity on the watermelon during its displacement from the roof to the ground

(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, $m=4.80\mathrm{kg},g=9.8{\mathrm{ms}}^{‐2}$and $s=18\mathrm{m}$

The work done will be

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

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

Determination of watermelon’s kinetic energy and speed before watermelon strikes the ground.

(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’s 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" $V^2={V^2}_0+2as$…..(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, $V_{1A}=0,g=9.8{\mathrm{ms}}^{‐2}$and $S=y_2-y_1=18\mathrm{m}$

Therefore,

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

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

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

Determination of watermelon’s kinetic energy and speed if there were air resistance

(c)

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

(a) The work done by gravity on watermelon during its displacement under free fall will remain unchanged since it does not depend on any other force.

(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.