Question

A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can's kinetic energy?

Answer 1

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 = $\frac{1}{2}$ I ω^2 + $\frac{1}{2} mv^{2}$

where

I = $\frac{1}{2} mr^{2}$ and ω = $\frac{v}{r}$

thus,

KE = $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ ($\frac{v}{r}$)^2 + $\frac{1}{2} mv^{2}$

     = $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ $\frac{v^{2} }{r^{2}}$ + $\frac{1}{2} mv^{2}$

     = $\frac{1}{4} mv^{2}$ + $\frac{1}{2} mv^{2}$

     = $\frac{3}{4} mv^{2}$

     = $\frac{3}{4} (0.46) (1.1)^{2}$

     = 0.42 J

Answer 2

The kinetic energy is mathematically given as

K.E=0.42J

What is the  kinetic energy?

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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