Question

An object, which is initially at rest on a frictionless horizontal surface, is acted upon by four constant forces. ?1F1 is 18.1 N18.1 Nacting due east, ?2F2 is 32.8 N32.8 N acting due north, ?3F3 is 47.8 N47.8 N acting due west, and ?4F4 is 10.8 N10.8 N acting due south. How much total work is done on the object in 3.33 s,3.33 s, if it has a mass of 22.0 kg22.0 kg?

total work done on object:

JJ

Which type of energy is changing for the object while the work is being done?

kinetic energy

internal energy

elastic potential energy

gravitational potential energy

How fast does the object end up moving at the end of the 3.33 s3.33 s?

final speed of object:

Answer #1

1) First you have to find the resultant of the forces:

R = (18.1 + 0 - 47.8 + 0)N i + (0 + 32.8 + 0 - 10.8)N j = -29.7 N i + 22 N j

acceleration:

a_x = -29.7N / 22kg = -1.35 m/s²

a_y = 22N / 22kg = 1 m/s²

displacement:

x = ½at² = ½ * -1.35m/s² * (3.33s)² = -7.49 m

y = ½ * 1m/s² * (3.33s)² = 5.54 m

work:

x-direction: Fx * x = -29.7N * -7.49m = 222.5 J

y-direction: Fy * y = 22N * 5.54m = 122 J

Total work = 222.5 + 122 = **344.5 J**

(work is scalar and has no direction)

2) KE

3) W = change in KE

=> 344.5 = ½*22*v^2

=> v= **5.6
m/s**

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