An object of mass 5 kg is dropped from a height of 10 m. Find it’s kinetic energy, when it is half way down.
Answers
Answered by
0
K. E =1/2 mgh
1/2*5kg*9.8(ms-2)*10m
=.....
1/2*5kg*9.8(ms-2)*10m
=.....
Answered by
2
u=0m/s
m= 5kg
h(to be covered)= 5m
KE= 1/2mv^2
[v^2=u^2+2as => v^2=2gs]
KE= 1/2 m (2gh)
[taking g=10m/s^2]
KE= 1/2 × 5 × 2 × 10 × 5
= 250 J
If u don't wanna give the whole solution then....according to conservation of energy, energy is equal at every point so energy at a distance of 5m will be equal to mgh
m= 5kg
h(to be covered)= 5m
KE= 1/2mv^2
[v^2=u^2+2as => v^2=2gs]
KE= 1/2 m (2gh)
[taking g=10m/s^2]
KE= 1/2 × 5 × 2 × 10 × 5
= 250 J
If u don't wanna give the whole solution then....according to conservation of energy, energy is equal at every point so energy at a distance of 5m will be equal to mgh
Similar questions