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aliina [53]
4 years ago
8

A ball having a weight of 1.5 N is dropped from a height of 4 meters. (Neglect air friction.) How much mechanical energy is "los

t" just before it hits the ground?
Physics
2 answers:
mafiozo [28]4 years ago
6 0
If there is no air resistance, then NO energy is 'lost'.

At the height of 4 meters above the ground, the ball has

                    (weight) x (height) 
                =    (1.5 N) x (4 m) 

                =         6 joules

of gravitational potential energy, relative to the ground.

If it's dropped, then the potential energy it has gets converted
to kinetic energy all the way down. 

Just before it hits the ground, it has no more potential energy,
but it has 6 joules of kinetic energy.

No energy is lost.  It just changes from potential to kinetic.
Both of them are forms of mechanical energy.
NikAS [45]4 years ago
4 0
As this mechanical energy is associated with height, it would be "Potential Energy" in particular.

U = mgh
U = F.h
U = 1.5 * 4
U = 6 Joules

So, 6 J of energy is lost before it hits the ground.

Hope this helps!
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1. An object is projected upward with a velocity of 125 m/s.
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Answer:

A) s = 796.38 m

B) t = 12.742 s

C) T = 25.484 s

Explanation:

A) First of all let's find the time it takes to get to maximum height using Newton's first equation of motion.

v = u + gt

u = 125 m/s

v = 0 m/s

g = 9.81 m/s²

Thus;

0 = 125 - 9.81(t)

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t = 125/9.81

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