A box of mass 4.5 kg is pushed across a rough surface (μK = 0.18) for a distance of 2.0 m by a constant force of 10 N. If the object reaches a speed of 2.0 m/s by the end of the push, what was its speed at the beginning of the push?

Explanation:

  Work done by force applied = force x displacement

= 10 x 2 = 20 J

Negative work done by frictional force

= μ mg x d where μ is coefficient of kinetic friction , m is mass and d is displacement

= - .18 x 4.5 x 9.8 x 2

= - 15.87 J

Net positive work done on the mass = 4.13 J

If v was the initial velocity

increase in kinetic energy = positive work done on mass

= 1 / 2 m v² - 1/2 m u² = 4.13 where v is final and u is initial velocity

1 /2 x 4.5 x 2 ² - 1/2 x 4.5 u² = 4.13

9 - 2.25 u² = 4.13

2.25 u² = 4.87

u² = 2.16

u = 1.47 m /s .

not too sure thinking all accelerations are pointed down