Question

Two blocks of masses of 2.00 kg and 4.00 kg are connected by a massless string going over a smooth, massless pulley. The table on which the smaller mass rest is frictionless. The other side of the 2.00 kg mass is connected to a spring of k=250 N/m and the far end of the spring is tied to a fixed point. The system is release from rest with the spring at its relaxed length.

A.) what is the speed of the hanging mass after it has fallen 10.0 cm? Show your reasoning.

B.) what is the K.E. Of the mass in the table at that instant? Show your work.

C.) what maximum distance does the hanging mass fall before coming to rest momentarily? Show your work.

Answer 1

As the hanging mass falls h =10 cm the spring also elongates by x =10 cm

Applying conservation of energy

0.0 = 0.5 k x² - Mgh + 0.5 (M+m) v²

Mgh = 0.5 (M+m) v² + 0.5 k x²

6 x 9.8 x 0.1 = 0.5 x 4 x v² + 0.5 x 250 x 0.1²

5.88 = 2v² + 1.25

v² = 2.315

v = 1.5215 m/s

The velocity of hanging mass after falling 10 cm is 1.5215 m/s

The kinetic energy of mass resting at table is = 0.5 m v² = 0.5 x 2 x 2.315 = 2.315 J

The maximum distance the mass M can fall before coming to rest v = 0.0 m/s

Again appying conservation of energy at that instant.

Since the heigth fallen is equal to the elongation of spring h=x

there for

Mgh = 0.5 k x² + 0.0

Mgh = 0.5 k h²

h = 2Mg/k = ( 2 x 4 x 9.8 ) / 250 = 0.3136 m = 31.36 cm

The maximum height fallen by the mass M before coming to rest is 31.36 cm