An initially stationary 1.3 kg object accelerates horizontally and uniformly to a speed of 9.4 m/s in 3.0 s. (a) In that 3.0 s interval, how much work is done on the object by the force accelerating it

57.434 Nm

Explanation: mass of object = 1.3kg, initial speed = 0m/s ( since it was at rest), final speed = 9.4 m/s.

Using the work energy theorem,

The work done on the object equals it kinetic energy.

Work done = mv²/2

Work done = 1.3 ×( 9.4 - 0)²/2

Work done = 1.3 × 9.4² / 2

Work done = 114.868 / 2 = 57.434 Nm

current flows in a wire

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