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
Question
Answer:
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