Two forces, Fand F2, act on the 7.00-kg block shown in the drawing. The coeffici

Question

Two forces, Fand F2, act on the 7.00-kg block shown in the drawing. The coefficient of kinetic friction ,-0.250. The magnitudes of the forces are F,-525 N and F2 = 33.0 N. The object moves to the east over a displacement s 5.00 m. Don't forget to include the weight (w), the normal force (Fx) and the kinetic friction (fi) a) Draw the free-body diagram of the system b) Determine the Net Work (total work) done over the displacement s. c) If the object started from rest, what is the final speed at the end of the displacement s. (tip: get the net force along to get F then fk k. F) 70.0 7.00

Explanation / Answer

Resolving force F1 towards horizontal, F1h= F1 cos(70) =525 cos(70)= 179.56 N

F1v = 525 sin(70) = 493.34 N

So net vertical force = F1v + W = Normal

N = 493.34+ 7*9.81 = 562 N

Friction = muk* N = 0.25*562 = 140.5 N

Net horizontal force towards East , Fh= 179.56-140.5-33= 6.06 N

Net work = Fh*s = 6.06*5 = 30.3 J ( ans)

0.5 mv^2 = 30.3 J

v= 2.94 m/sec (ans)

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