The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. If the ball was moving at 6.00m/s when the force was first applied, what is its final velocity?

a. We can calculate the amount of work by calculating the area under the graph.
first area (rectangular): 2.5 x 6 = 15
second area(trapezoid): 1/2 x (6+10) x 2.5 =20
total work done: 35 J
b. the force was first applied = 6 N
F = m.a
a = 6 : 3 = 2 m/s²
vf²=vi²+2as
vf²=6²+2.2.5
vf²=56
vf=7.5 m/s