An object, which is initially at rest on a frictionless horizontal surface, is a

Question

An object, which is initially at rest on a frictionless horizontal surface, is acted upon by four constant forces. F1 is 18.1 N acting due East, F2 is 41.2 N acting due North, F3 is 39.2 N acting due West, and F4 is 20.7 N acting due South. How much total work is done on the object in 2.22 seconds, if it has a mass of 20.0 kg? Number Units m/s Tools Which tyl × 102 s changing for the object while the above work is being done? O O O O internal energy kinetic energy elastic potential energy gravitational potential energy How fast does the object end up moving at the end of the 2.22 seconds? Number Units

Explanation / Answer

Ans)

Given forces F1=18.1N acting due east

F2=41.2N acting due north ,

F3=39.2N acting due west

F4=20.7N acting due south

For finding net force on the object is

Net north force F(n)=F2-F4=41.2N-20.7N=20.5N(north)

Net west force F(w)=F3-F1=39.2N-18.1N =21.1N=(west)

Resultant force F=sqrt[(F(n)2+F(w)2]

=sqrt[(20.5)2+(21.1)2]

F=29.41N

Find object acceleration

a=F/m

=(29.41)/20     (given m=20kg)

a=1.47 m/s2

find the distance when the object moves (starting from rest)

d=at2/2

d=[(1.47)x(2.22)2]/2

d=3.62m

find the work done on the object

W=Fxd

W=29.41x3.62

W=106.46J

V=at

V=1.47x2.22

V=3.26m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---