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 = 52.6 J and a maximum displacement from equilibrium of 0.277 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 = 0.277 m on a rough surface so that it loses 16.3 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)0.5kx^2=E

or 0.5*k*0.277^2=52.6

or k=1371.06 N/m

b)KE=52.6 J

c)0.5mv^2=E

or 0.5*m*3.45^2=52.6

or m=8.84 kg

d)conserving energy,

0.5*0.16^2*1371.06+0.5*8.84*v^2=52.6

or v=2.816 m/s

e) KE=0.5mv^2

=0.5*8.84*2.816^2

=35.05 J

f)potential energy=0.5kx^2

=0.5*1371.06*0.16^2

=17.55 J

g)0.5*1371.06*A^2=52.6-16.3

or A=0.23 m

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