You are testing a new amusement park roller coaster with an empty car with mass 127 kg. One part of the track is a vertical loop with radius 11.8 m. At the bottom of the loop (point A) the car has speed 27.1 m/s, and at the top of the loop (point B) it has speed 8.9 m/s. As the car rolls from point A to point B, how much work is done by the friction

Wf = 7876.5 J

Explanation:

To find the work done by the friction force you use the following formula, in which the total work is given by the change in the kinetic energy of the car:

 (1)

m: mass of the car = 127 kg

vB: speed at the top of the loop = 8.9 m/s

vA: speed at the bottom of the loop = 27.1 m/s

The total work is the contribution of the work done by the gravitational force and also the work done by the friction force, that is:

 (2)

g: gravitational acceleration = 9.8m/s^2

h: height of the car to the ground at point B = 2R = 2(11.8m) = 23.6 m

Then, you first equal the equations (1) and (2). Next, you replace the values of vB, vA, m, g and h, in order to obtain the work done by the friction force:

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.