A horizontal block-spring system with the block on a frictionless surface has to

Question

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 51.4 J and a maximum displacement from equilibrium of 0.206 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 = 0.206 m on a rough surface so that it loses 15.3 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant?
m

Explanation / Answer

a)   k = spring constant = 51.4*2/(0.206)^2 =2422.47 N/m

b) kinetic energy = 51.4 J

c) m = 51.4*2/(3.45)^2 = 8.63 kg

d) speed = [3.45^2 -0.16^2*2422.47/8.63]^0.5 =2.17 m/s

e) kinetic energy = 8.63*2.17^2/2 =20.3 j

f) potential energy =51.4 - 20.3 = 31.05 j

g) X = [ (51.4-15)*2/2422.7]^0.5 =0.17 m

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