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​

c. -288 J

Explanation:

Given;

mass of the block, m = 4 kg

velocity of the block, v = 12 m/s

deceleration of the block, a = 3 m/s²

Apply work-energy theorem;

work done in stopping the block = kinetic energy of the block

W = ¹/₂mv²

where;

W is the magnitude of the work done by the force

W = ¹/₂ x 4 x 12²

W = 288 J

Since the work done by the force is in opposite direction, then the value of the work done by the force is -288 J

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

568.98

explanation:

its math