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 = 47.7 J and a maximum displacement from equilibrium of 0.274m.

(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 = 0.274 m on a rough surface so that it loses 13.9 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 ,

a) let the spring constant is k

total mechanical eenergy = 0.5 * k * A^2

A is the amplitude

47.7 = 0.5 * k * 0.274^2

k = 1270.7 N/m

the spring constant is 1270.7 N/m

b)

at the equilibrium point ,

as the potential energy is zero

kinetic energy at the equilibrium point = total mechanical energy

kinetic energy at the equilibrium point = 47.7 J

c)

let the mass is m

for kinetic energy at the equilibrium point

0.5 * m * 3.45^2 = 47.7

m = 8.02 Kg

the mass of block is 8.02 Kg

d)
let the speed of block is v m/s

Using conservation of energy

0.5 * 1270.7 * 0.16^2 + 0.5 * 8.02 * v^2 = 47.7

solving for v

v = 2.8 m/s

the speed of the block is 2.8 m/s

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