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# A stone of mass 2 kg is dropped from a height 40 meter. Calculate the speed when it hits the ground. Assume a constant air resistance force 20 N acts on it.A. 8 m/sB. 10 m/sC. 12 m/sD. 16 m/sE. 0 m/s

Last updated date: 22nd Jun 2024
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Hint: Use Newton’s second law of motion to determine the net force acting on the stone to determine the acceleration produced in the stone. Then use the kinematic relation to determine the final velocity of the stone.

Formula used:
${v^2} = {u^2} + 2as$
Here, v is the final velocity, u is the initial velocity of the body, a is the acceleration, and s is the displacement.

Apply Newton’s second law of motion on a stone to determine the net acceleration produced in the stone as follows,
${F_{net}} = ma$
$\Rightarrow {F_{air}} - mg = ma$
Here, ${F_{air}}$ is the force due to air, m is the mass of the stone, g is the acceleration due to gravity and a is the net acceleration.
Substitute 20 N for ${F_{air}}$, 2 kg for m, and $10\,m/{s^2}$ for g in the above equation.
$20\,N - \left( {2\,kg} \right)\left( {10\,m/{s^2}} \right) = \left( {2\,kg} \right)a$
$\therefore a = 0\,m/{s^2}$
Now, we can use a kinematical equation relating final velocity, initial velocity, acceleration, and displacement to determine the final velocity of the stone.
${v^2} = {u^2} + 2as$
Here, v is the final velocity, u is the initial velocity of the stone, a is the acceleration, and s is the vertical
distance of the stone from its initial position to the ground.
Since the stone is dropped from a certain height, its initial velocity is zero.
Therefore, the above equation becomes,
${v^2} = 2as$
$\Rightarrow v = \sqrt {2as}$
Substitute $0\,m/{s^2}$ for a and 40 m for s in the above equation.
$v = \sqrt {2\left( {0\,m/{s^2}} \right)\left( {40\,m} \right)}$
$\therefore v = 0\,m/s$

So, the correct answer is “Option E”.

Note:
In this question, the net force on the stone is determined by the free body diagram of the forces acting on the stone. The force acting along the vertical upward direction is taken as positive whereas the force acting in the downward direction is taken as negative. Therefore, always specify the positive and negative directions of forces acting on the body.