Question

A horizontal block-spring system with the block on a
frictionless surface has total mechanical energy *E* = 41.4
J and a maximum displacement from equilibrium of 0.284 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* =
0.284 m on a rough surface so that it loses 13.3 J by the time it
reaches its first turning point (after passing equilibrium at
*x* = 0). What is its position at that instant?

m

Answer #1

here,

the total mechanical energy , ME = 41.4 J

the displacment from equilibrium , x = 0.284 m

a)

let the spring constant be K

ME = 0.5 * K * x^2

41.4 = 0.5 * K * 0.284^2

solving for K

K = 1026.6 N/m

b)

the kinetic energy of the system at the equilibrium point , KE = total mechanical energy = 41.4 J

c)

the maximum speed , v = 3.45 m/s

let the mass of block be m

KE = 0.5 * m * v^2

41.4 = 0.5 * m * 3.45^2

solving for m

m = 6.96 kg

d)

at x1 = 0.16 m

let the speed be v1

using conservation of momentum

0.5 * m * v1^2 + 0.5 * k * x1^2 = ME

0.5 * 6.96 * v1^2 + 0.5 * 1026.6 * 0.16^2 = 41.4

solving for v1

v1 = 2.85 m/s

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