Question

AT THE POSITION SHOWN IN THE DIAGRAM, THE MASS C m = 0.50 KG) HASKE 3.0 J AND usp = 2.0 J. THE SURFACE 15 FRICTYONLESS. 10.) FIND THE SPEED OF THE MASS AS IT PASSES THROUGH X0 (b) FIND THE SYSTEM AMPLITUDEL_(C) FIND THE SPRING'S ELASTIS CONSI

Answer 1

The Total Mechanical energy of the system at a particular point is,

Mechanical energy = kinetic energy + potential energy

now at x = 20 cm = 0.2 m,

Mechanical energy = 3 + 2 = 5 J

Now as the surface is frictionless so the system is an isolated system where energy remains conserved.

(a) Now when the mass passes the equilibrium position of x = 0 cm. the potential energy of spring becomes zero as it is directly proportional to the displacement from the equilibrium position which is zero in this case. therefore whole mechanical energy is converted into the kinetic energy of that mass(m = 0.5 kg)

Mechanical energy (x= 0 cm) = Kinetic energy of mass

5 = (1/2)*m*V²

5 = (1/2)*0.5*V²

solving for V,

V = 4.47 m/s

Velocity of the mass when it passes through x=0 cm is 4.47 m/s.

(c) at x= 20 cm = 0.2 m, The spring has potential energy of 2 Joules so,

(1/2)*k*x² = Usp

(1/2)*k*(0.2)² = 2

k = 100 N/m

Spring elastic constant is 100 N/m.

(b) Now when the mass is elongated maximum from its equilibrium position which is amplitude of the SHM. The whole mechanical energy is converted into potential energy of the spring.

Mechanical energy (x=A) = potential energy of the spring

5 = (1/2)*k*A²

5 = (1/2)*100*A²

A = 0.32 m

Amplitude of the system is 32 cm.