k = spring constant of the spring = 85 N/m
m = mass of the box sliding towards the spring = 3.5 kg
v = speed of box just before colliding with the spring = ?
x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m
the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.
Using conservation of energy
Kinetic energy of spring before collision = spring energy of spring after compression
(0.5) m v² = (0.5) k x²
m v² = k x²
inserting the values
(3.5 kg) v² = (85 N/m) (0.065 m)²
v = 0.32 m/s
The box will hit the spring with the speed of 0.320 m/s
Given data:
The spring constant is, [tex]k=85 \:\rm N/m[/tex].
The mass of box is, [tex]m=3.5 \;\rm kg[/tex].
The distance of compression is, [tex]x = 6.5 \;\rm cm =0.065\; m[/tex].
The kinetic energy of spring is converted into spring potential energy. Then,
Kinetic energy = spring potential energy
[tex]\dfrac{1}{2}mv^{2} = \dfrac{1}{2}kx^{2}\\v^{2} = \dfrac{kx^{2}}{m}\\v = \sqrt\dfrac{kx^{2}}{m}}\\\\v = \sqrt\dfrac{85 \times 0.065^{2}}{3.5}} \\\\v=0.320 \;\rm m/s[/tex]
Thus, the speed of box to hit the spring is 0.320 m/s.
Learn more about spring potential energy here:
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