An initially stationary 4.3 kg object accelerates horizontally and uniformly to a speed of 11 m/s in 3.4 s. (a) In that 3.4 s interval, how much work is done on the object by the force accelerating it

The work done on the object by the force accelerating it is 520.31 J

Explanation:

Given;

mass of an object = 4.3 kg

horizontal velocity of the object, v = 11 m/s

time of acceleration, t = 3.4 s

work done is given as the product of force and distance

Work done = Fd

horizontal distance traveled by the object within 3.4 s, is calculated as follows;

X = Vt + ¹/₂gt², gravity has little or no influence on horizontal displacement, thus g = 0

X = Vt

X = 11*3.4 = 37.4 m

Force on the object, F = ma = m(v/t) = 4.3(11/3.4) = 13.912 N

work done = Fd = 13.912 x 37.4 = 520.31 J

Therefore, the work done on the object by the force accelerating it is 520.31 J

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xd