Question

58. An object of mass 10 kg is dropped from a building of
height 40 m. Give the best description of relationship between
gravitational potential energy and kinetic energy after 2 sec.
(g = 10 m/s)​

Answer 1

Answer:

After 2 sec the K.E would equal to the P.E equal to 2000 J .

Explanation:

As we know at top the total energy of the body is in the form of potential energy and kinetic energy is zero

Total Energy = P.E (at top)

=> Total Energy = mgh = 10×10×40 = 4000 joule ........eq(1)

But as the mass is falling the P.E converts to K.E and at the ground surface the K.E is almost the total energy.

Now after 2 sec to find energy we have to find the new height first

using equation of motion

s = Vit + 1/2 at^2

on simplifying (putting a=g=10 and Vi=0)

$s = 5 {t}^{2} = 5(4) = 20m$

thus after two sec the object would be exactly at the mid of total height and hence its P.E woul equal to K.E, but to prive this, Lets find the P.E again and then the K.E

P.E(new) = mgh' = 10×10×20 = 2000 j

Now from eq(1) find the K.E as

K.E = T.E - P.E

=> K.E = 4000 - 2000 = 2000 j

thus proved***

Answer 2

Answer:

After 2 sec the K.E would equal to the P.E equal to 2000 J .

Explanation:

As we know at top the total energy of the body is in the form of potential energy and kinetic energy is zero

Total Energy = P.E (at top)

=> Total Energy = mgh = 10×10×40 = 4000 joule ........eq(1)

But as the mass is falling the P.E converts to K.E and at the ground surface the K.E is almost the total energy.

Now after 2 sec to find energy we have to find the new height first

using equation of motion

s = Vit + 1/2 at^2

on simplifying (putting a=g=10 and Vi=0)

thus after two sec the object would be exactly at the mid of total height and hence its P.E woul equal to K.E, but to prive this, Lets find the P.E again and then the K.E

P.E(new) = mgh' = 10×10×20 = 2000 j

Now from eq(1) find the K.E as

K.E = T.E - P.E

=> K.E = 4000 - 2000 = 2000 j

thus proved***

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Explanation: