An object of mass 5 kg is dropped from a certain height. Just before it strikes the ground it has a kinetic energy of 1250 J. From what height was the object dropped? Ignore air resistance and use g = 10 m/s2.
PE at start=KE at end
mgh=final KE
solve for h
a body is thrown up with a velocity of 20 m/s . calculate the maximum height attained by the body.
To find the height from which the object was dropped, we can use the concept of conservation of energy. The total mechanical energy of the object is conserved, which means it remains constant throughout its motion, neglecting any non-conservative forces like air resistance.
The mechanical energy of an object includes both its potential energy (PE) and kinetic energy (KE). At the start, when the object is dropped, it only has potential energy, and just before it strikes the ground, it only has kinetic energy.
The potential energy (PE) of an object is given by the equation: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
The kinetic energy (KE) of an object is given by the equation: KE = 0.5mv^2, where m is the mass of the object and v is its velocity.
Since we know the mass of the object (m = 5 kg), the final kinetic energy (KE = 1250 J), and the acceleration due to gravity (g = 10 m/s^2), we can set up the following equation using conservation of energy:
PE + KE = Constant
mgh + 0.5mv^2 = Constant
Substituting the known values, we get:
5gh + 0.5 * 5 * v^2 = Constant
Now, just before the object strikes the ground, its velocity (v) will be equal to zero. Therefore, we only consider the potential energy term (mgh):
5gh = Constant
1250 J = 5gh
Now, we can solve for h:
h = 1250 J / (5g)
h = 1250 J / (5 * 10 m/s^2)
h = 25 m
Therefore, the object was dropped from a height of 25 meters.