An object of mass 20kg is released from a height of 10m above the ground level. The kinetic energy of the object just before it hits the ground is
Answer Details
The kinetic energy of the object just before it hits the ground is 2000J. We can use the principle of conservation of energy to solve this problem. When the object is released from a height of 10m above the ground level, it has gravitational potential energy which is given by: PE = mgh where: m is the mass of the object g is the acceleration due to gravity h is the height of the object above the ground level Substituting the given values, we get: PE = (20 kg)(9.8 m/s^2)(10 m) = 1960 J As the object falls towards the ground, its potential energy is converted into kinetic energy. At the point just before it hits the ground, all of its potential energy will have been converted into kinetic energy. The kinetic energy is given by: KE = 1/2 mv^2 where: m is the mass of the object v is the velocity of the object To find the velocity of the object just before it hits the ground, we can use the principle of conservation of energy again. The total energy of the system (object + Earth) is conserved, so the potential energy at the top of the fall is equal to the sum of the kinetic and potential energies just before impact. Therefore, we can write: PE at the top = KE just before impact + PE just before impact We know that the PE at the top is equal to 1960 J, and the PE just before impact is zero (since the object is at ground level). Therefore: 1960 J = 1/2 (20 kg) v^2 Solving for v, we get: v = √(2(1960 J)/(20 kg)) = 20 m/s Substituting the value of v into the equation for kinetic energy, we get: KE = 1/2 (20 kg) (20 m/s)^2 = 2000 J Therefore, the kinetic energy of the object just before it hits the ground is 2000 J.