A 2kg object is released from rest 8m above the surface of the Earth, During the fall work done against air resistance is 60J. Just before it hits the surface its speed is: (Hint-take g=10 m/sec2)

Explanation

To solve the problem, we first calculate the gravitational potential energy (PE) of the object when it is released from a height of 8 m. The formula for potential energy is PE = mgh, where m = 2 kg, g = 10 m/s², and h = 8 m. Thus, PE = 2 kg * 10 m/s² * 8 m = 160 J.Next, we account for the work done against air resistance, which is given as 60 J. The effective energy available for conversion into kinetic energy (KE) just before hitting the ground is PE - work done against air resistance = 160 J - 60 J = 100 J. Using the kinetic energy formula KE = (1/2)mv², we can rearrange it to find v: v = sqrt((2 * KE) / m) = sqrt((2 * 100 J) / 2 kg) = sqrt(100) = 10 m/s.Thus, the final speed of the object just before it hits the ground is 10 m/sec. The other options (36 m/s, 40 m/s, and 45 m/s) do not correctly account for the work done against air resistance, leading to incorrect calculations.