Question

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 54.5 J and a maximum displacement from equilibrium of 0.285 m.

(a) What is the spring constant?
N/m

(b) What is the kinetic energy of the system at the equilibrium point?
J

(c) If the maximum speed of the block is 3.45 m/s, what is its mass?
kg

(d) What is the speed of the block when its displacement is 0.160 m?
m/s

(e) Find the kinetic energy of the block at x = 0.160 m.
J

(f) Find the potential energy stored in the spring when x = 0.160 m.
J

(g) Suppose the same system is released from rest at x = 0.285 m on a rough surface so that it loses 15.4 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant?
m

Answer 1

A) Given that E = 54.5 J

amplitude is A = 0.285 m

but E = 0.5*k*A^2 = 54.5

spring consatnt k = (54.5/(0.5*0.285^2)) = 1341.95 N/m

B)
KE = 54.5 J

C) 0.5*m*v^2 = 54.5

mass m = 54.5/(0.5*v^2) = 54.5/(0.5*3.45^2) = 9.15 kg

D) v = w*sqrt(A^2-x^2)

w = sqrt(k/m) = sqrt(1341.95/9.15) =12.11 rad/s

then v = 12.11*sqrt(0.285^2-0.16^2) = 2.85 m/s

E) KE = 0.5*m*v^2 = 0.5*9.15*2.85^2 = 37.16 J

F) PE = 0.5*k*x^2 = 0.5*1341.95*0.16^2 = 17.17 J

G) 0.5*k*x^2 = 54.5-15.4 = 39.1 J

x = sqrt(39.1/(0.5*1341.95)) = 0.241 m