Use Newton's second law to find the speed of the block as it hits the ground.?

Question

A 3.0 kg block is attached to a string that is wrapped around a 2.0 kg, 4.0-cm-diameter hollow cylinder that is free to rotate. The block is released 1.0 m above the ground.


Also need to:
Use conservation of energy to find the speed of the block as it hits the ground.

Answer 1

The gravity of this M kg block is Mg, which causes the accelaration of itself downward and causes the rotational accelaration of the cylinder (mass m):
Mg = Ma + ((mr^2)*a/r)/r = (M+m)a
thus a = Mg/(M+m)
Since the distance of the movement is 1m, we have;
0.5*at^2 = 1, or v^2 = (at)^2 = 2a = 2Mg/(M+m)
v = sqrt(2*3.0*9.8/5.0) = 3.4 (m/s)

Now use the energy conservation to solve this problem.
the initial potential energy: Mgh = Mg
The final kinetic energy: 0.5Mv^2 + 0.5I*w^2
= 0.5Mv^2 + 0.5(m*r^2)*(v/r)^2
= 0.5*(M+m)*v^2
Before M hits the ground, all the potential energy would be converted to kinetic energy:
Mg = 0.5*(M+m)*v^2
or v = sqrt(2*M*g/(M+m)) = 3.4 (m/s)

Answer 2

the swimmer pushes the water at the back of him and leaves a vacuum/area the place the water was once. as a result the water tries to get lower back there to even issues out - could be gravity, the great leveller is the stress it is at artwork here. so the water tries to fill the hollow - and fairly a number of the hollow is filled by potential of water that's at the back of the swimmer - as a result giving him a unfastened holiday forwards on a variety of wave.