A crate is subjected to a force of magnitude F = [150/s] N having a constant direction and where s is measured in meters.The mass of the crate is 2-kg and when s = 2 m, the crate is moving to the left with a speed of 4 m/s. Determine its speed when s = 6 m. The coefficient of kinetic friction between the crate and the ground is mk = 0.5.

1) 11.6 m/s
2) 14.5 m/s
3) 13.2 m/s
4) 10.5 m/s

Equation of motion: Since the crate slides, the friction force developed between the crate and its constant surface is
Ff = mk N = 0.5 N.
S Fy = may; N - [150/s] sin30o - 2(9.81) = 2(0)
N = [ (75/s) + 19.62 ] N

Principle of work and energy : The horizontal component of force F which acts in the direction of displacement does positive work.
The friction force :
Ff = 0.5[ (75/s) + 19.62 ] = [ (37.5/s) + 9.81 ] N which does negative work since its acts in the opposite direction to that of displacement.

The normal reaction N, the vertical component of force F and the weight of the crate do not displace hence do not work. Applying the conservation of energy principle we have:

T1 + S U1-2 = T2

1/2 (2) 42 +

So the speed of the crate when s = 6m is,
v = 11.6 m/s.