A 5kg ball is dropped from a height of 10m. Find a)the initial potential energy of the ball. b)the kinetic energy just before it reaches the ground c)calculate the velocity before it reaches the groud.
mass of ball(m)= 5kg
height(h)= 10m
acc. due to gravity (g)= 9.8m/s2
a) potential energy= mgh = 5 x 10 x 9.8=490J
b) by the law of conservation of energy we know that the decrease in potential energy at any point is seen as an equal amount of increase in kinetic energy.
therefore, the kinetic energy becomes 490 joules as potential energy becomes 0 just above the ground.( h becomes 0)( m x g x 0=0)
c) K.E.=1/2 mv2
490 = 1/2 x 5 x v2
v2 = 490 x 2/5= 196 m/s
v= 14 m/s
this was the solution hope u now understand it.
regards
height(h)= 10m
acc. due to gravity (g)= 9.8m/s2
a) potential energy= mgh = 5 x 10 x 9.8=490J
b) by the law of conservation of energy we know that the decrease in potential energy at any point is seen as an equal amount of increase in kinetic energy.
therefore, the kinetic energy becomes 490 joules as potential energy becomes 0 just above the ground.( h becomes 0)( m x g x 0=0)
c) K.E.=1/2 mv2
490 = 1/2 x 5 x v2
v2 = 490 x 2/5= 196 m/s
v= 14 m/s
this was the solution hope u now understand it.
regards