A loose nut from a bolt on the bottom of an elevator which is moving up the shaft at$3\dfrac{m}{s}$ falls freely. The nut strikes the bottom of the shaft in 2s. Distance of the elevator from the bottom of the shaft when the nut fell off is:
A. 19.6 m.
B. 13.6 m.
C. 9.8 m.
D. 3.8 m.
Answer
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Hint:-The elevator is moving upwards and the nut is falling in the downwards therefore the time taken by the nut to fell on the bottom of shaft will be less than the situation in which the elevator is not moving at all as the nut will have some initial velocity in upwards direction.
Formula used: The formula of the Newton’s law of motion is given by $s = ut + \dfrac{1}{2}a{t^2}$ where u is the initial velocity t is the time taken and a is the acceleration of the body.
Complete step-by-step solutionIt is given in the problem that a nut from the elevator falls from the bottom of the elevator to the shaft when the elevator is moving upwards with a velocity of $3\dfrac{m}{s}$ the fall of the nut is free fall and the time taken for this is 2s and we need to find the distance which the nut travels.
As the relation in the Newton’s law of motion is given by,
$s = ut + \dfrac{1}{2}a{t^2}$
Where u is the initial velocity t is the time taken and a is the acceleration of the body.
As the acceleration is acceleration due to gravity and the initial velocity of the nut is in upwards direction therefore we get.
$ \Rightarrow s = \left( { - u} \right) \cdot t\dfrac{1}{2}g{t^2}$
Replace the values of initial velocity, the time taken and the acceleration due to gravity in the above relation.
$ \Rightarrow s = \left( { - 3 \times 2} \right) + \dfrac{1}{2}\left( {9 \cdot 8} \right){\left( 2 \right)^2}$
$ \Rightarrow s = \left( { - 6} \right) + \left( {9 \cdot 8} \right)\left( 2 \right)$
$ \Rightarrow s = 13 \cdot 6m$
The distance travelled by the nut is equal to $s = 13 \cdot 6m$. The correct answer for this problem is option B.
Note:- The time taken by the nut to fall off on the bottom of the shaft is less as there is movement of the elevator and that too in the upwards direction. If the elevator would be moving in the downwards direction then the time taken would have been as usual.
Formula used: The formula of the Newton’s law of motion is given by $s = ut + \dfrac{1}{2}a{t^2}$ where u is the initial velocity t is the time taken and a is the acceleration of the body.
Complete step-by-step solutionIt is given in the problem that a nut from the elevator falls from the bottom of the elevator to the shaft when the elevator is moving upwards with a velocity of $3\dfrac{m}{s}$ the fall of the nut is free fall and the time taken for this is 2s and we need to find the distance which the nut travels.
As the relation in the Newton’s law of motion is given by,
$s = ut + \dfrac{1}{2}a{t^2}$
Where u is the initial velocity t is the time taken and a is the acceleration of the body.
As the acceleration is acceleration due to gravity and the initial velocity of the nut is in upwards direction therefore we get.
$ \Rightarrow s = \left( { - u} \right) \cdot t\dfrac{1}{2}g{t^2}$
Replace the values of initial velocity, the time taken and the acceleration due to gravity in the above relation.
$ \Rightarrow s = \left( { - 3 \times 2} \right) + \dfrac{1}{2}\left( {9 \cdot 8} \right){\left( 2 \right)^2}$
$ \Rightarrow s = \left( { - 6} \right) + \left( {9 \cdot 8} \right)\left( 2 \right)$
$ \Rightarrow s = 13 \cdot 6m$
The distance travelled by the nut is equal to $s = 13 \cdot 6m$. The correct answer for this problem is option B.
Note:- The time taken by the nut to fall off on the bottom of the shaft is less as there is movement of the elevator and that too in the upwards direction. If the elevator would be moving in the downwards direction then the time taken would have been as usual.
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