A Mountain bar has a mass of 6.7×10−2 kg and a calorie rating of 270 cal.

What speed would this candy bar have if its kinetic energy were equal to its metabolic energy? [Note: The nutritional calorie, 1 Cal , is equivalent to 1000 calories (1000 Cal ) as defined in physics. In addition, the conversion factor between calories and joules is as follows: 1 = 1000 Cal = 1KCal = 4186 J .]

To find the speed of the candy bar, we first need to calculate its kinetic energy and metabolic energy.

The kinetic energy (KE) of an object can be calculated using the formula:

KE = (1/2) * mass * speed^2

The metabolic energy (ME) is given as 270 cal. To convert it to joules, we'll use the conversion factor provided:

1 cal = 4186 J

1 Cal = 1000 cal

Therefore, 270 cal = (270/1) * (4186/1) J = 270 * 4186 J

Now, let's equate the kinetic energy and metabolic energy:

KE = ME

(1/2) * mass * speed^2 = ME

Substituting the given values, we get:

(1/2) * (6.7×10^(-2)) kg * speed^2 = 270 * 4186 J

Simplifying the equation:

(6.7×10^(-2) * speed^2) = (270 * 4186 * 2)

Dividing both sides of the equation by (6.7×10^(-2)), we get:

speed^2 = (270 * 4186 * 2) / (6.7×10^(-2))

Now, we can solve for speed by taking the square root of both sides of the equation:

speed = √[(270 * 4186 * 2) / (6.7×10^(-2))]

Calculating the value:

speed ≈ √[2277480 / (6.7×10^(-2))]

speed ≈ √[2277480 * (10^2 / 6.7)]

speed ≈ √[2277480 * 14.92]

speed ≈ √[33970957.6]

speed ≈ 5828.8 m/s

Therefore, the speed of the candy bar would be approximately 5828.8 meters per second if its kinetic energy were equal to its metabolic energy.