Question

A block of mass 0.25 kg is against a spring compressed at 0.20 m with spring constant 50 N/m. When the spring is released, the block moves along the frictionless surface until entering a region with the coefficient of kinetic friction equal to 0.30 (when the block enters the friction region it is no longer in contact with the spring ). How far,L,into the region with friction does the block slide before stopping?

Answer #1

Mass of block = m = 0.25 kg

Spring constant = k = 50 N/m

Initial compression of spring = X = 0.2 m

Coefficient of kinetic friction on rough floor = = 0.3

Speed of block after it leaves contact with the spring =
V_{1}

Distance travelled by the block on the rough floor = L

Final speed of the block = V_{2} = 0 m/s (Block comes to
a stop)

The potential energy in the spring is converted into kinetic energy of the block.

kX^{2}/2 = mV_{1}^{2}/2

(50)(0.2^{2})/2 = 0.25(V_{1}^{2})/2

V_{1} = 2.828 m/s

From the free body diagram,

N = mg

f = N

f = mg

ma = f

ma = mg

a = g

a = (0.3)(9.81)

a = 2.943 m/s^{2}

a = -2.943 m/s^{2} (Negative sign as it is
deceleration)

V_{2}^{2} = V_{1}^{2} + 2aL

0^{2} = 2.828^{2} + 2(-2.943)L

L = 1.36 m

The block slides 1.36m into the region with friction before stopping.

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