Question

7.
A ball (100 gm) thrown from ground with speed
20 m/s comes back to the ground with speed
10 m/s. Find the work done by air resistance.
(1) 15 J
(2) 20 J
(3) 30 J
(4) 40 J​

Answer 1

The work done by air resistance is (1) 15 J

Step-by-step explanation:

This question uses the Work-Energy theorem of physics.

Given:

• mass $m = 100 \hspace{ 1mm} g= 0.1 \hspace{1 mm} kg$
• Initial velocity $v_1= -20 \hspace {1 mm} m/s$ (upwards)
• Final velocity $v_2= 10 \hspace {1 mm} m/s$ (downwards)

To be found:

The work done by air resistance $W_{air}$

Concepts used:

• Work Energy theorem states that the change in kinetic energy of a body is equal to the total work done on the body by all external forces.
• So, $Change\hspace{1 mm} in \hspace{1 mm} KE= \frac{1}{2}m(v_2^2-v_1^2)= total \hspace{1 mm} W$

We can see that in this question, two forces does work on the body:

• Gravity: gravity does net zero work on the body since the direction of gravity is constant all throughout upward and downward motion
• Air resistance: since air resistance always acts opposite to velocity direction, so $W_{air}$ is non-zero.

We can put the values in the work-energy theorem formula,

$\frac{1}{2}m(v_2^2-v_1^2)= \frac{1}{2}*0.1*(10^2-(-20)^2)= total \hspace{1 mm} W= 0+W_{air}$

We can find the value of $W_{air}$ by simplifying this equation,

⇒$W_{air}=\frac{0.1}{2}*(100-400)=-15 \hspace {1 mm} J$

So, The work done by air resistance is (1) 15 J