a body falls freely undef gravity .its speed is v when it lost an amount U of the gravitational energy .then its mass is
Answers
This problem depends on the principle of conservation of mechanical energy. That means, if an object is moved, the mechanical energy should be constant in every position.
Mechanical energy = Constant
Kinetic Energy + Potential Energy = Constant
(1/2)mv^2 + mgh = Constant
Now we can solve this problem by using above theorem.
As shown in attached diagram we can take 2 positions.
m = mass of the body
h = starting height of the mass reference to the potential energy zero level
u = starting velocity
v = velocity at 2nd position
g = gravitational acceleration
x = falling distance
by using principle of conservation of mechanical energy,
mechanical energy at 1st position = mechanical energy at 2nd position
(1/2)mu^2 + mgh = (1/2)mv^2 + mg(h-x)
0 + mgh = (1/2)mv^2 + mgh - mgx
mgx = (1/2)mv^2
but, lost of potential(gravitational) energy = mgx = U
Therefore U = (1/2)mv^2
mass of the body = m = 2U/(v^2)
