A crate having a mass of 200 kg, is subjected to two forces. If it is originally at rest, determine the distance it slides in order to attain a speed of 3 m/s. The coefficient of kinetic friction between the crate and the surface is m_k = 0.25.

1) 2.34 m
2) 3.50 m
3) 4.45 m
4) 5.35 m

Equation of motion: Since the crate slides, the friction force developed between the crate and its constant surface is
F_f = m_k N = 0.25 N
S F_y = ma_y
N - 200 (9.81) + 500 (3/5) = 200 (0)
N = 1962 - 300 = 1662 N

Principle of work and energy : The horizontal component of force 400 N and 500 N which act in the direction of displacement does positive work.
The friction force :
F_f = 0.25 (1662) = 415.5 N does the negative work since its acts in the opposite direction to that of displacement.

The normal reaction N, the vertical component of force 500 N and the weight of the crate do not displace hence do not work.Since the crate is originally at rest T₁ = 0. Applying the conservation of energy principle we have:
T₁ + S U_1-2 = T₂
400 cos0^o(s) + 500 (4/5)(s) - 372.4(s) = 1/2(200) (3²)
where s is the distance travelled by the crate
400s + 400s - 415.5 s = 900
s = 2.34 m

So the distance it slides in order to attain a speed of
3 m/s is s= 2.34 m.