Dynamics_Part28

# Dynamics_Part28 - 1 2 mv 2 2 − μ 2 mg f δ − 1 2 k δ...

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Denoting the box’s velocity at the bottom of the chute by v 2 ,wehave 1 2 mV 2 +2 mg f w sin θ 1 2 cos θ W = 1 2 mv 2 2 On the horizontal surface, the only tangential force acting until the box encounters the spring is friction, which is μ 2 mg to the left. The spring force is F e = kx and the work done in compressing the spring a distance δ is 1 2 k δ 2 . Thus, the work done as the box moves from the beginning of the horizontal surface until it comes to rest is U 2 3 = μ 2 mg ( f + δ ) 1 2 k δ 2 The Principle of Work and Energy tells us that T 2 + U 2 3 = T 3 Now, since the box comes to rest, its speed is v 3 =0 , so that its kinetic energy is T 3 =0 . Thus,
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Unformatted text preview: 1 2 mv 2 2 − μ 2 mg ( f + δ ) − 1 2 k δ 2 = 0 Substituting for 1 2 mv 2 2 from above, this equation becomes 1 2 mV 2 + 2 mg f w sin θ − 1 2 cos θ W = μ 2 mg ( f + δ ) + 1 2 k δ 2 We are given δ = 1 2 f and μ 2 = 1 3 , wherefore μ 2 mg ( f + δ ) + 1 2 k δ 2 = 1 3 mg w 3 2 f W + 1 2 k w 1 2 f W 2 = 1 2 mg f + 1 8 k f 2 Combining these two equations yields 1 2 mg f + 1 8 k f 2 = 1 2 mV 2 + 2 mg f w sin θ − 1 2 cos θ W Solving for k , we conclude that k = 4 mg f w 4 sin θ − 2 cos θ − 1 + V 2 g f W...
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