Two constant forces act on an object of mass m = 4.20 kg object moving in the xy

Question

Two constant forces act on an object of mass m = 4.20 kg object moving in the xy plane as shown in the figure below. Force F with arrow1 is 20.0 N at 35.0°, and force F with arrow2 is 49.0 N at 150°. At time t = 0, the object is at the origin and has velocity (3.50î + 3.00) m/s. (a) Express the two forces in unit-vector notation. F with arrow1 = 16.38 i +11.47 j Correct: Your answer is correct. . N F with arrow2 = 42.43 i +24.5 j Correct: Your answer is correct. . N (b) Find the total force exerted on the object. 26.05 i +35.97j Correct: Your answer is correct. . N (c) Find the object's acceleration. 6.20i + 8.56 j Correct: Your answer is correct. . m/s2 Now, consider the instant t = 3.00 s. (d) Find the object's velocity. 18.6 i+25.68 j Incorrect: Your answer is incorrect. . m/s (e) Find its position. c m (f) Find its kinetic energy from ½mvf2. kJ (g) Find its kinetic energy from ½mvi2 + F with arrow · vector r. kJ (h) What conclusion can you draw by comparing the answers to parts (f) and (g)?

Explanation / Answer

Object velocity = Vi +at = (3.50î + 3.00) +( 18.6 i+25.68 j ) = (-15.1 i^ + 28.68 j^) m/sec (ans)

Object position = 0.5at^2 = ( -27.9 i ^ + 38.52 j^) m (ans)

KE = 0.5*4.2*1050.55 = 2206.16 J (ans)

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