HW_13_sol - Solutions HW 13 P112 S07 1) m = 0.150 kg . k =...

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Solutions HW 13 P112 S07 1) kg 150 . 0 = m . N/m 300 = k . When m 0120 . 0 1 = x , m/s 300 . 0 1 = v . Because the spring force is conservative, mechanical energy K U E + = is conserved. (Assume that no friction force acts on the toy.) (a) 2 2 1 2 2 1 2 1 2 1 2 1 2 1 ) m/s 300 . 0 )( kg 150 . 0 ( ) m 0120 . 0 )( N/m 300 ( + = + = + = mv kx K U E J 10 84 . 2 2 ! " = E . (b) When the toy is at its maximum displacement from the equilibrium position it will be momentarily at rest, so there 0 = K and all of the mechanical energy is potential energy: 2 2 1 max kA U E = = . m 10 37 . 1 N/m 300 ) J 10 84 . 2 ( 2 2 2 2 ! ! " = " = = k E A . (c) The toy will have its maximum speed whenever 0 = U , at the equilibrium position: 2 max 2 1 max mv K E = = . m/s 615 . 0 kg 150 . 0 ) J 10 84 . 2 ( 2 2 2 max = ! = = " m E v . 2) The maximum acceleration of both blocks, assuming that the top block does not slip, is a max = kA ( m + M ), and so the maximum force on the top block is m m + M ( ) kA = μ s mg , and so the maximum amplitude is A max = s ( m + M ) g k . 3) φ = π /2, φ = −π /4, φ = 0, , φ = π /4, φ = π /2 4) Let’s assume the car weighs 1000 kg, the person 980-N, and the center of gravity sinks about 3 cm as the person climbs in the car. We’ll model the situation
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HW_13_sol - Solutions HW 13 P112 S07 1) m = 0.150 kg . k =...

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