A rope of negligible mass can support a load of M kg. Prove that the mass of the greatest load which can be raised is equal to(M )/(1+ (2h)/(g t^2))kg, where g is the acceleration due to gravity andh is the height through which the said load rises from rest with uniform acceleration in time t.

A rope of negligible mass can support a load of M kg. Prove that the m

1.	A rope of negligible mass can support a load of M kg. Prove that the mass of the greatest load which can be raised is equal to(M )/(1+ (2h)/(g t^2))kg, where g is the acceleration due to gravity andh is the height through which the said load rises from rest with uniform acceleration in time t.
Answer» SOLUTION :Since the rope can support a load of M kg, the maximum tension the rope can withstand is given by T = Mg (figure a). Now, if a mass m is raised by the rope with a uniform ACCELERATION a as shown in figure (b), then the net Upward force on the mass m = (T" - mg) where T. is the tension in the string. Then, from Newton.s second law of motion, ` T. = mg =ma ORT. =m(G+a)` FORTHE greatestvalueof m ` T.=T thereforem (g +a) = Mg ` ` thereforem ( Mg)/(g+a) = (M )/(1 +a//g)` Now, from the equation of motion, `s=ut + 1/2 at^2` Weget,`h= 0+1/2a t^2or a = (2h)/(t^2)` fromequation(i )`m=(M )/(1 +(2h)/(g t^2) ) kg .`