Respuesta :
Answer
given,
mass of steel ball, M = 4.3 kg
length of the chord, L = 6.5 m
mass of the block, m = 4.3 Kg
coefficient of friction, μ = 0.9
acceleration due to gravity, g = 9.81 m/s²
here the potential energy of the bob is converted into kinetic energy
[tex]m g L = \dfrac{1}{2} mv^2[/tex]
[tex]v= \sqrt{2gL}[/tex]
[tex]v= \sqrt{2\times 9.8\times 6.5}[/tex]
v = 11.29 m/s
As the collision is elastic the velocity of the block is same as that of bob.
now,
work done by the friction force = kinetic energy of the block
[tex]f . d = \dfrac{1}{2} mv^2[/tex]
[tex]\mu m g. d = \dfrac{1}{2} mv^2[/tex]
[tex]d=\dfrac{v^2}{2\mu g}[/tex]
[tex]d=\dfrac{11.29^2}{2\times 0.9 \times 9.8}[/tex]
d = 7.23 m
the distance traveled by the block will be equal to 7.23 m.
