A body of mass 100kg is released from a height of 200m. With what energy does the body strike the ground (g =10m/s2)
with what energy does the body strike the grounds level?
KE = PE = mgh = 100*10*200 = 200,000J
PE=MGH
mass=100kg,gravity=10m/s,height=2m.
100*10*2=2000joules
To determine the energy with which the body strikes the ground, we need to consider two types of energy: potential energy and kinetic energy.
1. Potential Energy: The potential energy of an object at a height is given by the formula PE = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the mass (m) is 100 kg, the acceleration due to gravity (g) is 10 m/s^2, and the height (h) is 200 m. Therefore, the potential energy of the body is:
PE = 100 kg * 10 m/s^2 * 200 m = 200,000 J
2. Kinetic Energy: The kinetic energy of an object in motion is given by the formula KE = 0.5 * m * v^2, where m is the mass of the object, and v is its velocity. In this case, the body is initially at rest, so its velocity is 0 when it is released. However, when the body reaches the ground, it will have a velocity determined by the fall time and the acceleration due to gravity. The fall time can be calculated using the formula t = sqrt((2h)/g), where h is the height and g is the acceleration due to gravity. In this case:
t = sqrt((2 * 200 m) / (10 m/s^2)) = sqrt(40 s^2) ≈ 6.32 s
Now, we can calculate the velocity using the formula v = g * t:
v = 10 m/s^2 * 6.32 s ≈ 63.2 m/s
Finally, we can calculate the kinetic energy of the body at the moment it strikes the ground:
KE = 0.5 * 100 kg * (63.2 m/s)^2 ≈ 200,320 J
Therefore, the body strikes the ground with approximately 200,320 Joules of energy.