A 5kg block on a rough horizontal surface is attached to a light spring (force constant=1.6kN/m). the block passes through its equilibrium position with a kinetic energy of 5J and is brought momentarily to rest after stretching the spring .06m. how much work is done by the frictional force on the block as it moves from its equilibrium position to the point of momentary rest?

2.12 J

Explanation:

Initial kinetic energy = final elastic energy + work by friction

KE = EE + W

KE = ½ kx² + W

5 J = ½ (1600 N/m) (0.06 m)² + W

W = 2.12 J

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

it is given that trillions of microbes make up to 1% to 3% of the body mass.

let's take average of this percentage: 2%

consider the average mass of the human body as 70 kg.

then, trillions of microbes would have mass  

therefore, 100 trillion micro-organisms would weigh:

this is an approximate value and could change depending on the mass of the human body and percentage of the microbes present in it.

average mass per microorganism =