Question

A metal ball of mass 60 g falls on a concrete floor from a vertical height of 2.8m and rebounds to a height of 1.3 m. Find the change in K.E. in S.I. units.

Answer 1

When a body moves vertical direction under the gravity , then Change in kinetic energy + change in potential energy = 0
e.g., ∆K.E + ∆P.E = 0
So, here no require to find out velocity of body in different points. Just find change in potential energy , change in kinetic automatically we can get by using above relation.

Now come to the question,
A ball falls from a height 2.8m , after striking ball rebounds 1.3m above the ground . So, separation between final position to intial position of particle(h) is = 1.3 - 2.8 = -1.5 m [ here negative sign shows that body appears below the reference level]
Now, change in potential energy = mg× change in height
= mg × speration between final position to initial position( h )
Here m = 0.06 Kg , g = 10 m/s² and h = -1.5 m
So, change in potential energy = - 0.06 × 10 × 1.5 = 0.9J

So, change in kinetic energy + change in potential energy = 0
Change in kinetic energy -0.9J = 0
change in kinetic energy = 0.9J

Answer 2

Given,

Initial Height = 2.8 m

Final Height = 1.3 m

mass = 60 g = 60/1000 Kg = 0.06 Kg

Acceleration due to Gravity = 10 m/s^2

As energy remains Constant every where.


=> P.E = mg( Final Height - Initial Height)

=> P.E = 0.06 x 10 x ( 1.3 - 2.8)

=> P. E = 0.6 x (-1.5) Joules

=> P.E = -0.9 Joules


As, Law of Conservation of Energy

=> P.E + K.E = 0

=> 0.9 J + K.E = 0

=> K.E = 0.9 Joules


Law of Conservation states that energy can neither be created nor be destroyed. In other words, we can say that energy remains Constant at every palce.





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