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 mk = 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 Ff = mk N = 0.25 N S Fy = may 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 : Ff = 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 U1-2 = T2 400 cos0o(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.