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The 10 Kg ball is dropped from a height of 10m. Find
(a) the initial potential energy of the ball
(b) the kinetic energy just before it reaches the ground, and
(c) the velocity just before it reaches the ground.
A) (a) $580$ ,
     (b) $980\,J$
     (c) $18\,m/s$
B) (a) $920$ ,
     (b) $880\,J$
     (c) $14\,m/s$
C) (a) $950$ ,
     (b) $880\,J$
     (c) $12\,m/s$
D) (a) $980$ ,
     (b) $980\,J$
     (c) $14\,m/s$

seo-qna
Last updated date: 23rd May 2024
Total views: 46.2k
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Answer
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46.2k+ views
Hint: Use the formula of the potential energy, substitute all known values to find the answer. Use the law of conservation of energy to find the kinetic energy from potential energy. Substitute the kinetic energy value in its formula to find the velocity formula.

Formula used:
(1) The potential energy is given by
$PE = mgh$
Where $PE$ is the potential energy of the ball, $m$ is the mass of the ball, $g$ is the acceleration due to gravity and $h$ is the height at which the object is placed from the earth surface.

(b) The formula of the kinetic energy is
$KE = \dfrac{1}{2}m{V^2}$
Where $KE$ is the kinetic energy while the ball falling down, $m$ is the mass if the ball and $V$ is the velocity of the ball.

Complete step by step solution:
It is given that the
Mass of the ball, $m = 10\,Kg$

Height at which the ball is dropped, $h = 10\,m$

(a) By using the formula of the potential energy,
$PE = mgh$
Substitute the value of the acceleration due to gravity as $9.81$ .
$PE = 10 \times 10 \times 9.81$
By the further simplification of the above step,
$PE = 980\,J$

(b) According to the law of the conservation of the energy, the total energy always remains constant. That is the amount of energy before falling is equal to the total energy after falling.
$PE = KE$
Hence $KE = 980\,J$

(c) The formula of the kinetic energy is
$KE = \dfrac{1}{2}m{V^2}$
Substitute the values of the mass and the kinetic energy in it.
$980 = \dfrac{1}{2} \times 10 \times {V^2}$
By simplifying the above equation,
${V^2} = \dfrac{{980 \times 2}}{{10}}$
By further simplification,
${V^2} = 196\,m{s^{ - 1}}$
$V = 14\,m{s^{ - 1}}$

Thus, the option (D) is correct.

Note: In this question, the potential energy of the ball is calculated before the ball is dropped. The kinetic energy of the ball is calculated when the ball drops down with some kinetic energy in it. These both are equal in this case, based on laws of conservation of energy.