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 body's situation won't change the forces are equal opposit and on the same body

terd muffin is the correct answer