Question

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?

Answer 1

Answer:

Explanation:

Given

mass of box$=2.5 kg$

speed $v=1.2 m/s$

compression in the spring $x=2 cm$

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

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

Thermal energy generated is $E=0.5 J$

this thermal energy is produced by friction force

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

$f_r=\mu _k\times N$

where $\mu_k =coefficient\ of\ kinetic\ friction$

$N=mg$

$f_r=\mu _kmg$

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

$0.5=\mu _k\cdot 0.5$

$\mu _k=1$

$\mu _k=1$ contradicts the fact that friction is present because $\mu _k=1$ shows there is no friction so either surface is friction less or compression in the system should be more than 2 cm