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 = 43.7 J and a maximum displacement from equilibrium of 0.227 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 = 0.227 m on a rough surface so that it loses 14.6 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

We know total energy of a spring

E = (1/2)*k*A2

=> 43.7 = 0.5*k*0.2272

=> k = 1696.13 N/m

(a) Potential energy of the system = (1/2)*k*x2

= 0.5*1696.13*0.1602 = 21.71 J

Kinetic energy = E - 21.71 = 21.99 22 J

(b) Energy stored in spring = P.E. = 21.71 J 21.7 J

(c) at start it has energy = (1/2)*k*x2 = 43.7 J

In which 14.6 J is lost

Energy left = 43.7 -14.6 = 29.1 J

Now since at tirning point v = 0 i.e. K.E. = 0

So (1/2)*k*y2 = 29.1 => y = 0.185 m

SO another extreme point (turniing point) => x = -y = -0.185 m

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