A 2.00-kg object is observed at t = 0 to be at the origin and moving in the + x

Question

A 2.00-kg object is observed at t = 0 to be at the origin and moving in the + x direction at 9.00 m/s. You can assume that there are no forces acting along the x-direction. At t = 0 a constant, uniform force of +3.00 j N (in the y-direction) is applied to the object. a) Determine the position r (with its x and y components) and velocity v (with its x and y components) of the object at t = 5.0 s. Remember that you can treat the x and y motion independently and then combine them in the end to express kinematic parameters. r(5.0 s) = _______________i + ______________j v (5.0 s)= ______________ i + ______________j b) How much work was done on the object by the applied force during this 5.0 s interval? Work = _____________________________________________ c) Use the definition of kinetic energy (i.e. evaluate ½ mv2) to determine the kinetic energy at t = 0 and at t = 5.0 s. KE (0) = ___________________________; KE(5.0 s) = __________________________ d) What is the change in KE during this 5-s interval? Use the values from c) above. Change in KE = _____________________________________________

Explanation / Answer

m= 2 kg, Ux = 9 m/s, Fy = 3 N,

a) ay ( accleration along y axis0 = 3/2 = 1.5 m/s^2 j

R ( after 5 seconds) = 45 i + 18.75 j ( X= vt , y = 1/2at^2)

V (affter 5 seconds ) = 9 i + 1.5 x 5 j = 9 i + 7.5 j m/s

b)W = ( 0 i + 5j) ( 45 i + 18.75 j) = 93.75 J

c)I nitial KE = 1/2 (2) (9)^2= 81 J

Velocity at 5 seconds = sqroot ( 9^2 + 7.5^2) = 11.715 m/s

final KE = 1/2 (2) ( 11.715)^2=137.25 J

change in KE = 56.25 J

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