Question

e4) The acceleration due to gravity on the earth's surface is a =-980665k meters per second squared. For this problem, let's approximate this as a =-10k m/(sec2). This means that gravity acting on a 5 kg object exerts a force of F ma 50k N, where N, a Newton, is a unit of force equal to a kg·m/ (see) If I raise a 5 kg object up 7 meters, I must do 350 J of work, where J, a Joule, is a unit of work (or energy) equal to a N-m. (If I lower the object 7 meters, I'm doing -350 J of work.) If a climber wants to pull up to herself a 50 meter long dangling rope whose density is 3 kg/m, how much work must she do?

Answer 1

This can easily be solved by the method of center of gravity(CG), considering it as a point mass

for a uniform object like this rope, it's CG lies on midway along its length,

now

W= mgh

where, m= 50*3 = 150 kg, g= 10 m/s², h= 50/2 = 25 m

W=37500 J