
A ball of 1 kg is released from the tower of Pisa. The kinetic energy generated in it after falling through 10 m will be?
A. 10 J
B. 9.8 J
C. 0.98 J
D. 98 J
Answer
585.3k+ views
Hint: Basic laws of motion are to be applied here. The acceleration due to gravity g is to be substituted in place of a third law of motion which will give the final velocity and kinetic energy arises due to motion (or velocity) of a body.
Kinetic energy:
$K.E. = \dfrac{1}{2}mv^2 $
Third law of motion:
$v^2 - u^2 = 2as$
Complete step by step answer:
When objects freely fall under the influence of gravity they experience and acceleration of g = 9.8 m/$s^2$ .
No matter what the mass of the object is, it will reach the ground at the same time with the same velocity. This velocity can be obtained from third law of motion as:
$v^2 = 2gs$
Since the object was at rest before, the initial velocity will be zero (u=0). We require to know the velocity of the object after a distance of 10 m so s = 10 m and g = 9.8 m/$s^2$ gives us:
$v^2 = 2 \times 10 \times 9.8 m^2/s^2$
Let us leave this intact as we will require it in the kinetic energy expression. The kinetic energy of the ball of mass 1 kg after dropping a distance of 10 m vertically becomes:
$K.E. = \dfrac{1}{2} \times 10 kg \times 2 \times 10 \times 9.8 m^2/s^2$
Simplification of this gives us just:
$K.E = 98 kg m^2/s^2$ .
The standard unit of kinetic energy is Joules so we get 98 J as the required kinetic energy. Therefore the correct answer is option (D).
Additional information:
If you drop a feather and a heavy ball at the same time from the same height it would reach the ground at the same time given that there is no air resistance. Air resistance is responsible to slow the feather down as it reaches the ground therefore we practically observe that the feather reaches the ground later.
Note:
Even if one uses the value of g to be 10 m/$s^2$ , one will obtain the result as 100 J which will be very close to our answer 98 J. Also, it is better to leave the $v^2$ obtained as it is. Finding out the root in the expression will create unnecessary calculations further.
Kinetic energy:
$K.E. = \dfrac{1}{2}mv^2 $
Third law of motion:
$v^2 - u^2 = 2as$
Complete step by step answer:
When objects freely fall under the influence of gravity they experience and acceleration of g = 9.8 m/$s^2$ .
No matter what the mass of the object is, it will reach the ground at the same time with the same velocity. This velocity can be obtained from third law of motion as:
$v^2 = 2gs$
Since the object was at rest before, the initial velocity will be zero (u=0). We require to know the velocity of the object after a distance of 10 m so s = 10 m and g = 9.8 m/$s^2$ gives us:
$v^2 = 2 \times 10 \times 9.8 m^2/s^2$
Let us leave this intact as we will require it in the kinetic energy expression. The kinetic energy of the ball of mass 1 kg after dropping a distance of 10 m vertically becomes:
$K.E. = \dfrac{1}{2} \times 10 kg \times 2 \times 10 \times 9.8 m^2/s^2$
Simplification of this gives us just:
$K.E = 98 kg m^2/s^2$ .
The standard unit of kinetic energy is Joules so we get 98 J as the required kinetic energy. Therefore the correct answer is option (D).
Additional information:
If you drop a feather and a heavy ball at the same time from the same height it would reach the ground at the same time given that there is no air resistance. Air resistance is responsible to slow the feather down as it reaches the ground therefore we practically observe that the feather reaches the ground later.
Note:
Even if one uses the value of g to be 10 m/$s^2$ , one will obtain the result as 100 J which will be very close to our answer 98 J. Also, it is better to leave the $v^2$ obtained as it is. Finding out the root in the expression will create unnecessary calculations further.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

