A spring (k = 600 N/m) is at the bottom of a frictionless plane that makes an angle of 30° with the horizontal

Question

A spring (k = 600 N/m) is at the bottom of a frictionless plane that makes an angle of 30° with the horizontal. The upper end of the spring is depressed 0.10 m, and a 2.0-kg block is placed against the depressed spring. The system is then released from rest. What is the kinetic energy of the block at the instant it has traveled 0.10 m and the spring has returned to its uncompressed length?

2.0 J

1.8 J

2.2 J

1.6 J

1.0 J

Answer 1

♦ initial stock of energy of the system is

E=0.5*k*x^2, where k=600N/m, x=0.1m;

It is wholly the stock of the spring;

♥ final stock of energy is

E=E1+E2, where E1 is kinetic energy of the block,

E2=mgh is pot energy, h=x*sin30° is altitude at the instant in question;  

♠ E=E; 0.5*k*x^2 = E1 +mgh, hence

E1= 0.5*k*x^2 –mg*x*sin30 = 2.02 J;