A body of mass 5kg falls from a height of 10m above the ground. What is the kinetic energy of the body just before it strikes the ground? (Neglect energy lo...

The potential energy of an object is given by mgh, where m is the mass of the object, g is the acceleration due to gravity and h is the height of the object. As the object falls from a height of 10m, its potential energy is converted to kinetic energy, which is given by (1/2)mv², where m is the mass of the object and v is its velocity. At the instant just before the object strikes the ground, all its potential energy has been converted to kinetic energy. Therefore, the kinetic energy just before the object strikes the ground is given by: KE = (1/2)mv² where m = 5kg and g = 10ms^-2 The velocity of the object just before it strikes the ground can be found using the equation: v² - u² = 2gh where u is the initial velocity of the object (which is zero, as the object is dropped from rest). Substituting the given values, we get: v² = 2gh = 2 x 10 x 10 = 200 v = sqrt(200) = 14.14ms^-1 Substituting the value of v in the equation for kinetic energy, we get: KE = (1/2)mv² = (1/2) x 5 x 14.14² = 500J Therefore, the kinetic energy of the body just before it strikes the ground is 500J. Answer: 500J