An object of mass 2.5 kg has speed of 15 m/s when height is 5m above ground what is speed when 12 m above ground?
We can solve this problem by using the principle of conservation of mechanical energy. The mechanical energy of an object is the sum of its kinetic energy (KE) and potential energy (PE), given by the equation:
E = KE + PE
The potential energy of an object near the surface of the Earth can be calculated using the formula:
PE = m * g * h
where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s² near the surface of the Earth), and h is the height above the reference point (in this case, the ground).
Initially, when the object is 5 m above the ground, its potential energy is:
PE1 = m * g * h1
where h1 = 5 m.
The initial kinetic energy can be calculated using the formula:
KE1 = 0.5 * m * v1²
where v1 is the initial velocity of the object. In this case, v1 = 15 m/s.
The initial mechanical energy is the sum of the initial kinetic energy and potential energy:
E1 = KE1 + PE1
Now, when the object is 12 m above the ground, its potential energy is:
PE2 = m * g * h2 = m * g * 12
To find the final velocity of the object (v2) when it is 12 m above the ground, we need to equate the initial mechanical energy (E1) to the final mechanical energy (E2), assuming there is no energy loss due to friction or air resistance:
E1 = E2
KE1 + PE1 = KE2 + PE2
0.5 * m * v1² + m * g * h1 = 0.5 * m * v2² + m * g * h2
Now we can solve this equation for v2:
0.5 * m * v1² + m * g * h1 - m * g * h2 = 0.5 * m * v2²
v2² = (0.5 * m * v1² + m * g * h1 - m * g * h2) / (0.5 * m)
v2 = √[(0.5 * m * v1² + m * g * h1 - m * g * h2) / (0.5 * m)]
v2 = √[v1² + 2 * g * (h1 - h2)]
Substituting the values given in the question, we can calculate the speed when the object is 12 m above the ground.