A 4 kg block is moving at 12 m/s on a horizontal frictionless surface. a constant force is applied such that the block slows with an acceleration of 3 m/s^2. how much work must this force do to stop the block? a. -576 J
b. -360 J
c. -288 J
d. 360 J
e. 576 J​

The block comes to a rest from 12 m/s with acceleration 3 m/s^2, which is carried out over a distance x such that

(12 m/s)^2 - 0^2 = 2 (3 m/s^2) x

=>   x = (12 m/s)^2 / (2 (3 m/s^2)) = 24 m

The force itself has a magnitude F such that

F = (4 kg) (3 m/s^2)

=>   F = 12 N

This force is pointing opposite the direction in which the block is moving, so the work it's performing is negative, and the work done is

W = - (12 N) (24 m) = -288 Nm = -288 J

which makes C the answer.

the second one and the forth one are in the purple because they are opinions and the rest are in the yellow

explanation:

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.