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 = 56.3 J and a maximum displacement from equilibrium of 0.262 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 = 0.262 m on a rough surface so that it loses 16.5 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

Here,

total mechnical energy , E = 56.3 J

maximum displacement , A = 0.262 m

a) let the spring constant is k

E = 0.5 * k * A^2

56.3 = 0.5 * k * 0.262^2

k = 1640 N/m

b) at the equilibrium position

kinetic energy of the system = total energy of system

kinetic energy of the system = 1640 N/m

c)

let the mass is m

0.5 *m ( v^2) = kinetic energy

0.5 * m * 3.45^2 = 56.3

m = 9.46 Kg

the mass of block is 9.46 Kg

d)

at x = 0.160 m

let the speed of block is v

Using conservation of energy

0.5 * 9.46 * v^2 + 0.5 * 1640 * 0.160^2 = 56.3

solving for v

v = 2.73 m/s

the speed of the block is 2.73 m/s

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