A boy has a bag of sand of mass 20 kg . First of all he keeps the bag on his head and moves 10 m
Answers
So, total work done by man: -
F × S cos ¢(theta)
20 × 10 cos 90
As cos 90= 0
So, F×S cos ¢=0
Hence total work done by man is ZERO.
Work done in horizontal direction = 0 as the potential energy is not changed and the change in kinetic energies : initial and final - is zero.
work done when the person carries the suitcase in vertical direction =
= m g h = change in potential energy = 30 kg * 9.8 m/sec/sec * 10 m
= 2940 Joules
=============
There is a force F1 lifting the body upwards. There is Weight W = m g pulling the body downwards. The net force F2 on the body is upwards with acceleration a.
F2 = m a = F1 - W = F1 - m g
a = F1 / m - g
Work done by Force F1 with which the object is lifted = F1 . s = 230 Joules
F1 = energy spent / distance = 230 J / 2 m = 115 Newtons
Net Acceleration of the body = a = F1 / m - g = 115/10 - 10 = 1.5 m/sec/sec
==========================
m = 2 kg F = 8 Newtons t = 12 sec
a = F/m = 4 m/sec/sec
v = u + a t = 0 + 4 * 12 = 48 m/sec
kinetic energy = 1/2 m v² = 1/2 * 2 * 48 * 48 = 48² Joules