
A coconut of mass $m$ falls from its tree through a vertical distance of $s$ and could reach ground with a velocity of $v$$m{s^{ - 1}}$ due to air resistance. The work done by air resistance is:
A. $ - \dfrac{m}{2}(2gs - {v^2})$
B.$ - \dfrac{1}{2}m{v^2}$
C. $ - mgs$
D. $m{v^2} + 2mgs$
Answer
585.3k+ views
Hint:In this question,we are going to apply the concept of work done.Work done by air resistance can be calculated by comparing it with the work done by gravity and actual work done by air.
Complete step by step answer:
It is given in the question that,
mass of coconut$ = m.$
Distance travelled vertically by the coconut = s
Velocity =v
Now, the work done by the coconut due to gravity be ${W_g} = mgs$ . . . (1)
$\therefore $change in velocity of coconut is $ = \dfrac{1}{2}m{(u = v)^2} = - \dfrac{1}{2}m{(v)^2}$ . . . (2)
Now, we have the total work done on the coconut is equal to change in kinetic energy, hence,
${W_a} + {W_g} = \dfrac{1}{2}m{v^2}$
Where, ${W_a} \to $work done by air resistance and from (1) and (2)we get,
${W_a} = \dfrac{1}{2}m{v^2} - {W_g}$
${W_a} = \dfrac{1}{2}m{v^2} - mgs$
$\therefore {W_a} = - \dfrac{m}{2}(2gs - {v^2})$
Hence, the correct option is option (A) $ - \dfrac{m}{2}(2gs - {v^2})$
Note: The total change in velocity of the coconut is to be considered by taking the value of initial and final velocity of the coconut. Also remember that work done can be positive, negative and zero. It depends on the angle between the applied force and displacement of the body.
Complete step by step answer:
It is given in the question that,
mass of coconut$ = m.$
Distance travelled vertically by the coconut = s
Velocity =v
Now, the work done by the coconut due to gravity be ${W_g} = mgs$ . . . (1)
$\therefore $change in velocity of coconut is $ = \dfrac{1}{2}m{(u = v)^2} = - \dfrac{1}{2}m{(v)^2}$ . . . (2)
Now, we have the total work done on the coconut is equal to change in kinetic energy, hence,
${W_a} + {W_g} = \dfrac{1}{2}m{v^2}$
Where, ${W_a} \to $work done by air resistance and from (1) and (2)we get,
${W_a} = \dfrac{1}{2}m{v^2} - {W_g}$
${W_a} = \dfrac{1}{2}m{v^2} - mgs$
$\therefore {W_a} = - \dfrac{m}{2}(2gs - {v^2})$
Hence, the correct option is option (A) $ - \dfrac{m}{2}(2gs - {v^2})$
Note: The total change in velocity of the coconut is to be considered by taking the value of initial and final velocity of the coconut. Also remember that work done can be positive, negative and zero. It depends on the angle between the applied force and displacement of the body.
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