Physics, asked by divyanshsharma17abd, 10 months ago

A body of mass 30 kg is thrown vertically upward with initial speed of 30 m/s. Its total energy after 3 s will be

Answers

Answered by rahulsingh10a
6

Answer:

13.5 kj (approx)

Explanation:

E = 1/2 mv^2

Answered by amitnrw
7

Total energy after 3 seconds will be​ 13.5 kJoule if A body of mass 30 kg is thrown vertically upward with initial speed of 30 m/s

Explanation:

Energy can neither be created nor destroyed.

it can be converted from one form to another form

Hence Energy at 0sec  & 3 sec will be same

Energy at secs

= PE + KE

= mgh  + (1/2)mv²  

= 0  +  (1/2)(30)(30)²

= 13500  Joule

= 13.5 kJoule

initially Potential Energy = 0

Kinetic energy = 13.5 kJoule

Total Energy = 13.5 kJoule

Energy at 3 Sec will also be 13.5 kJoule

Verification :

g = 10 m/s²

V = U + at

a = -g = - 10  m/s²

t = 3 sec

Velocity after 3 secs = 30 - (10)(3) = 0

Height after 3 sec  

S = (V² - U²)/2a

=> Height  = (0² - 30²)/2(10) = 45 m

Energy at 3 secs

= mgh + (1/2)mv²  

= 30(10)(45)  + 0

=   13500  Joule

= 13.5 kJoule

Potential Energy at 3 secs = 13.5 kJoule

Kinetic energy at 3 secs = 0

Total Energy = 13.5 kJoule

13.5 kJoule Kinetic energy converted into potential energy

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