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Talja
12 days ago
9

A 2.5-kg box, sliding on a rough horizontal surface, has a speed of 1.2 m/s when it makes contact with a spring. The block comes

to a momentary halt when the compression of the spring is 2.0 cm. The thermal energy generated by friction, from the instant the block makes contact with the spring until it comes to a momentary halt, is 0.5 J (b) What is the coefficient of kinetic friction between the box and the rough surface?
Physics
1 answer:
ValentinkaMS [3.4K]12 days ago
4 0

Answer:

Explanation:

Given

The box's mass=2.5 kg

speed v=1.2 m/s

compression in the spring x=2 cm

Total Initial energy K.E.=\frac{mv^2}{2}

K.E.=\frac{2.5\times 1.2^2}{2}=1.8 J

The thermal energy generated is E=0.5 J

which is produced by the friction force

The Work done by the Friction Force is W=f_r\cdot s

f_r=\mu _k\times N

This contradicts the presence of friction since \mu_k =coefficient\ of\ kinetic\ frictionsuggests no friction, indicating either a frictionless surface or that the compression in the system exceeds 2 cm

N=mg

f_r=\mu _kmg

0.5=\mu _k\times 2.5\times 10\times 0.02

0.5=\mu _k\cdot 0.5

\mu _k=1

\mu _k=1

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