.A stone of mass 2kg is dropped from rest, reaches a velocity of 8 ms-1 just before reaching the ground. What is the loss in its potential energy ?
Answers
Answer:
Explanation:
Given,
u=3m/s
v=0
a=g=10m/s
2
(velocity of a body moving upwqard is decreasing so acceleration is negative)
v
2
=u
2
+2as
0=3
2
+2×−10×s
0=9−20s
20s=9, s=
20
9
U=mgh=20×10×
20
9
=90J
Given : .A stone of mass 2kg is dropped from rest, reaches a velocity of 8 ms-1 just before reaching the ground.
To Find : the loss in its potential energy ?
Solution:
Energy can neither be created nor destroyed it can be converted from one form to another form.
Total Energy = KE + PE
KE = kinetic Energy
PE = Potential Energy
loss in its potential energy = Gain in Kinetic energy
KE = (1/2)mv²
PE = mgh
Dropped from rest hence
initial KE = (1/2) * 2(0)² = 0
reaches a velocity of 8 m/s
Hence
Final KE = (1/2)*2 (8)² = 64
Gain in KE = 64 - 0 = 64 Joule
Loss in PE = 64 Joule
loss in its potential energy 64 Joule
Another method :
u can use v² - u² = 2as
a = g
to find initial height and final height = 0
and use mgh formula
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