Energy created by falling object

AI Thread Summary
A 1.36 kg weight falling 0.1524 meters generates energy through the conversion of gravitational potential energy to kinetic energy. The calculation for this energy involves using the formula mgΔh, where m is mass, g is the acceleration due to gravity, and Δh is the height fallen. The expected energy output is approximately 2.032 joules, which raises questions about the power generated compared to the object's static state. It's important to note that a falling object does not create more power than it has in potential energy at rest. Understanding these energy transformations is crucial in physics.
Greg E
Messages
1
Reaction score
0
If there is a 1.36 kg weight falling .1524 meters,
how many joules are created when the object reachs the end of the .1524 meters. How do you calculate this?
I came up with 2.032 joules which is .20723 kg meters which I think is enough power to lift 1.36 kg .1524 meters which is the same as where I started and that cannot be correct because a falling object has more power than the same object standing still.
Thanks,
Greg
 
Physics news on Phys.org
This seems like a homework problem from Introductory Physics.

However, an object falling implies falling due to gravity. What one has is a conversion/transformation from gravitational potential energy to kinetic energy.

Taking the acceleration of gravity to be constant, the change in gravitational potential energy is mg\Delta h, where \Delta h is the change in elevation.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top