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 = 38.7 J and a maximum displacement from equilibrium of 0.253 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 = 0.253 m on a rough surface so that it loses 12.9 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) E = 1/2 k A^2
k = 2*38.7/0.253^2= 1209.2

b) at equilibrium, PE = 0
so KE = E = 38.7 J

c)
KE = 1/2 mv^2
m = 2*38.7/3.45^2= 6.5 kg

d)
0.5*6.5*v^2=38.7 - 0.5*1209.2*0.16^2
v=2.67 m/s

e)KE + PE = E

KE = 38.7 - 0.5*1209.2*0.16^2 = 23.22 J

f)PE = 0.5*1209.2*0.16^2= 15.48 J

g)

Ei + Wfriction = Ef

0.5*1209.2*0.253^2 - 12.9 = 0.5*1209.2*x^2
x=-0.207 m

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