Question

Besides the gravitational force, a 2.50-kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of (5.40î − 3.30ĵ) m, where the direction of ĵ is the upward vertical direction. Determine the other force.

Answer 1

Initial speed u = 0

time t = 1.20 s

Mass m = 2.50 Kg

Displacement S = 5.40 i - 3.30 j   m

From Kinematic relation

                       S = ut + (1/2) at^2

  5.40 i - 3.30 j = 0 + 0.5a(1.20)^2

  5.40 i - 3.30 j = 0.72 a

Acceleration a = (5.40 i - 3.30 j )/(0.72)

                       = 7.2 i - 4.583 j   m/s^2

Net force F = ma

                  = (2.50 Kg)(7.2 i - 4.583 j)

                = 18 i - 11.45 j N  

One force is gravitatinal force F1 = - mg j

                                                    = - (2.50*9.8) j

                                                    = - 24.5 j N

Let F2 be the another force, then

                    F2 = F - F1

                         = (18 i - 11.45 j) - (-24.5 j)

                         = 18 i - 11.45 j+ 24.5 j

                         = 18 i + 13.05 j N

Magnitude |F2| = Sqrt[18^2 + 13.05^2]

= 22.232 N

Thanks, Have a nice day

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