Calculating Work and Power in Two Common Problems: Bubble Radius & Heart Pumping

[mprm86]

1. There's an spheric bubble in the air with a radius of 3.2 cm. You blow air inside the bubble until its radius gets 5.8 cm. Which is the work done on the bubble? Consider the (constant) superficial tension of the bubble as 26 N/m.

2. The heart pumps 75 ml of blood in each beating with an average pressure of 110 mmHg. It beats 65 times per minute. Calculate the power of pumping of the heart.

 

[HallsofIvy]

How about actually reading the "sticky" thread titled "read this before posting"? Is that too much to ask?

 

[quicknote]

1. To calculate the work done on the bubble, we can use the formula W = ∫Fdx, where W represents work, F is the force applied, and dx is the displacement. In this case, the force is due to the change in surface tension of the bubble, which is given as 26 N/m. The displacement is the change in radius, which is 5.8 cm - 3.2 cm = 2.6 cm. Converting to meters, we get 0.026 m.

Plugging these values into the formula, we get W = 26 N/m * 0.026 m = 0.676 Nm. Therefore, the work done on the bubble is 0.676 Nm.

2. To calculate the power of pumping of the heart, we can use the formula P = W/t, where P represents power, W is the work done, and t is the time taken. In this case, the work done is the product of force and displacement, which is the product of pressure and volume of blood pumped. Converting the given values to SI units, we get 75 ml = 0.075 L and 110 mmHg = 14.665 kPa.

Plugging these values into the formula, we get P = (14.665 kPa * 0.075 L)/60 s = 0.0183 kW. Therefore, the power of pumping of the heart is 0.0183 kW or 18.3 W.