Question

A constant 75-N force acts on a 6.00 kg box. The box initially is moving at 20 m/s in the direction of the force, and 3.0 s seconds later the box is moving at 58 m/s. Determine both the work done by this force and the average power delievered by the force during the 3.0-second interval.

Answer 1

Work done = 8752 J

Power = 2917 W

Step-by-step explanation:

Mass of the box, m= 6.00 kg

Force , F = 75 N

Initial velocity , vi = 20 m/s

The velocity after t = 3.0 s is vf = 58 m/s

Acceleration of the box is ,
`(v - u)/(t)`

a = (58m/s-20m/s)/3s

a = 12.7 m/s²

The distance moved by the box is

v² - vi² = 2as

(58)² -(20)^2 = 2*12.7 m/s^2 * s

s = 116.69 m

Work done, W = F.s

= 75 N * 116.69 m

= 8752 J

Power = W/t

= 8752 J/3s

P = 2917 W

Answer 2

Answer: 299,427.84 J, 99,809.28 W

Explanation: we will assume the acceleration of the body to be constant, hence newton's laws of motion is applicable.

From the question

Force (f) = 75 N

Mass of object (m) = 6kg

Initial velocity (u) = 20 m/s

Final velocity (v) = 58 m/s

Time taken (t) = 3s

Recall that

F = ma

75 = 6a

a = 75/6 = 12. 5 m/s²

We need to get the distance before we can get the work done.

Recall that v² = u² + 2as, where s = distance covered

56² = 20² + 2(12.5)s

56² - 20² = 25s

2736 = 25s

s = 2736/ 25 = 109.44m

Work done = force × distance

Work done = 75 × 109.44 = 299,427.84 j

Power = work done / time

Power = 299,427.84 / 3

Power = 99,809.28 W