Question

Part A (Work Energy) Problem Al: A force F = 24N acts at 60° above the horizontal on a 3kg block that is attached to a spring whose stiffness constant is k = 200. Assume uk = 0.1. The system starts at rest with the spring unextended. If the block moves 40cm, find the work done 24N k = 20N/m 60° N 3kg 1) By the force F. 2) By the friction 3) By the spring 4) What is the final speed of the block? Page 1 of 4

Answer 1

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+ 24singo 24N - 24 cos 60° F-24 N 6 = 60 m = 3kg elk = 0.1 k = 2ON-m x = 40cm = 0.4 m. UKN SON Balancing vertical forces, N + 24 sin 60 = 30 N = 9.215 N Frictional force = MKN = 0.189.215 = 0.9215 N Spring Force = kx = 20x0.4 = 8 N Work done by F = Fx cose = 24 x 0.4x COS 60° = 485 ✓ Work done by Friction = UKNX x = 0.9215 x0.4 = 0.3686 J ✓ Work done by spring = K x ² = 2 x 20 x0.42 = 3.2JV Balancing Horizontal forces, Fcoso - kx - MxN=ma 24 cos 60° - 20x0.4 -0.1 X 9.215 = 3xa a = 1.026 m/s2 v=u²+2ax 0.4 - 0 + 2 x 1.026 V = 0.906 m/s