Question

m1 = 2.2 kg block slides on a frictionless horizontal surface and is connected on one side to a spring (k = 45 N/m) as shown in the figure above. The other side is connected to the block m2 = 4 kg that hangs vertically. The system starts from rest with the spring unextended.

a) What is the maximum extension of the spring?
m

a) What is the speed of block m2 when the extension is 45 cm?

Answer 1

a) Total mass = (2.2+4)= 6.2kg

The mass falls is distance x.

Potential energy lost by going down is m*g*x=4*g*x (eq 1)

The potential energy gained by spring is 1/2*k*x²  (eq 2)

So equating eq 1 and eq 3 we get : 4*g*x=1/2*k*x

So, x =4*g*2/(k) = 1.74 m.

b)As the mass falls the distance x that is 1.74m, nothing will move as the spring will reach its maximum spring extension x.At x the speed should be 0 that is when the force of gravity is equal to the force of spring.