A25.0 kg pickle is accelerated from rest through a distance of 6.0 m in 4.0 s across a level floor. if the friction force between the pickle and the floor is 3.8 n, what is the work done to move the object?

Accelerating forwards; fa > ff            fn                           |ff< · > fa                      |                     fgfn = fg. so ∑f is just equal to sum of fa and ff (fn and fg cancel)find the distance the pickle travels. v₀ = 0 m/sd = 6.0 mt = 4.0 suse kinematic equationd = v₀t + at²/2d = 0t + at²/2d = at²/2a = 2d/t²a = 2(6.0 m)/(4.0 s)² a = 0.75 m/s² forwards∑f = fa + ffwe are given that ff = 3.8 n backwardsfa = ∑f - fffa = ma - ff fa = (25.0 kg)(0.75 m/s²) - (-3.8n)fa = 22.55 n forwardssow = fa · dw = (22.55 n)(6.0 m)w = 135 jw = 140 jthe work done to move the object is 140 j.