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

Answer: W=260.174J

Explanation: since the object is stationary, it initial velocity U = 0

final velocity V = 11 m/s, time t = 3.4s, distance S = ?, acceleration a = ? work done W = ? force F = ?

W = FS

a = v-u/t = 11-0/3.4 = 3.2353m/s^2

to calculate the distance, let look at one of the equations of motion

V^2=U^2+2as hence s = V^2-U^2/2a = 11*11/2*3.235 = 18.7017m

But force F = MA (mass*acceleration)

F = 4.3*3.2353 = 13.91179N

therefore work done W = 13.91179*18.7017 = 260.174J