A 5kg mass is dropped from a height of 30m.wheni it is 18m above the ground (g=10mls²)
To find the velocity of the 5kg mass when it is 18m above the ground, you can use the principle of conservation of mechanical energy.
The potential energy at a height h is given by the formula PE = m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height. The initial potential energy of the mass is PE_initial = m * g * 30m.
According to the conservation of mechanical energy, the total energy (potential energy + kinetic energy) of the object remains constant throughout its motion. So, initially, only potential energy exists, and when the mass is 18m above the ground, it will be partially converted into kinetic energy.
The final potential energy of the mass at a height of 18m is PE_final = m * g * 18m.
So, the initial potential energy (PE_initial) minus the final potential energy (PE_final) will be equal to the kinetic energy (KE) of the mass at that point:
PE_initial - PE_final = KE
Plugging in the values, we get:
(5kg * 10m/s² * 30m) - (5kg * 10m/s² * 18m) = KE
(1500kg⋅m²/s²) - (900kg⋅m²/s²) = KE
600kg⋅m²/s² = KE
The kinetic energy is given by the formula KE = (1/2) * m * v², where v is the velocity of the mass.
So, we can set up the equation:
600kg⋅m²/s² = (1/2) * 5kg * v²
Simplifying:
600kg⋅m²/s² = 2.5kg * v²
v² = (600kg⋅m²/s²) / (2.5kg)
v² = 240m²/s²
Taking the square root of both sides:
v = √(240m²/s²)
v ≈ 15.49 m/s
Therefore, the velocity of the 5kg mass when it is 18m above the ground is approximately 15.49 m/s.