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 = 42.4 J and a maximum displacement from equilibrium of 0.288 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 = 0.288 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?

Explanation / Answer

Here,

E = 42.4 J

A = 0.288 m

a) let the spring constant is k

E = 0.50 * k * A^2

42.4 = 0.50 * k * 0.288^2

k = 1022 N/m

the spring constant is 1022 N/m

b)

as all the energy will be stored as kinetic energy

the kinetic energy of the system at the equilibrium point is 42.4 J

c)

let the mass is m

42.4 = 0.50 * m * 3.45^2

m = 7.12 Kg

the mass is 7.12 Kg

d)

let the speed is v

0.50 * 1022 * 0.160^2 + 0.5 * 7.12 * v^2 = 42.4

v = 2.87 m/s

the speed of the mass is 2.87 m/s

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