Question

An object of mass 6.50 kg undergoes a displacement S= 11.5 m i + 5.5 m j while it is being acted on by a net force F = 5.9 N i -2.2 N j. a) How much work is done on the object by the net force? b) If the object above starts from rest, what is its speed after it has undergone the displacement described above?

Answer 1

A)

Work = Force.displacement

Dot product of Force and displacement
Work = (11.5, 5.5) . ( 5.9, - 2.2) =(11.5*5.9) - (5.5*2.2)= 67.85 - 12.1 = 55.75 J

B)

Using the work energy theorem , we find the speed of the object; the work energy theorem states that the change in kinetic energy of the object equals the work done, so if it starts from rest, its final KE equals work done.

1/2 m v^2 = 50.55J or v = 4.29m/s

m = 6.5 kg

$\frac{1}{2} (6.5) * v^2 = 55.75$

v = 4.14 m/s is the is its speed of the object after it has undergone the displacement described above.