Answer:
the can's kinetic energy is 0.42 J
Explanation:
given information:
Mass, m = 460 g = 0.46 kg
diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m
velocity, v = 1.1 m/s
the kinetic energy of the can is the total of kinetic energy of the translation and rotational.
KE = [tex]\frac{1}{2}[/tex] I ω^2 + [tex]\frac{1}{2} mv^{2}[/tex]
where
I = [tex]\frac{1}{2} mr^{2}[/tex] and ω = [tex]\frac{v}{r}[/tex]
thus,
KE = [tex]\frac{1}{2}[/tex] [tex]\frac{1}{2} mr^{2}[/tex] ([tex]\frac{v}{r}[/tex])^2 + [tex]\frac{1}{2} mv^{2}[/tex]
= [tex]\frac{1}{2}[/tex] [tex]\frac{1}{2} mr^{2}[/tex] [tex]\frac{v^{2} }{r^{2}} [/tex] + [tex]\frac{1}{2} mv^{2}[/tex]
= [tex]\frac{1}{4} mv^{2}[/tex] + [tex]\frac{1}{2} mv^{2}[/tex]
= [tex]\frac{3}{4} mv^{2}[/tex]
= [tex]\frac{3}{4} (0.46) (1.1)^{2}[/tex]
= 0.42 J
The kinetic energy is mathematically given as
K.E=0.42J
Question Parameter(s):
A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s .
Generally, the equation for the Kinetic energy is mathematically given as
KE =0.5I w^2 + 0.5mv^{2}
Therefore
KE = 0.5*0.5 mr^{2} (v/r)^2 + 0.5mv^{2}
K.E= 3/4 mv^{2}
K.E= 3/4 (0.46) (1.1)^{2}
K.E=0.42J
In conclusion, the can's kinetic energy
K.E=0.42J
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