A 5-lbm toy car is at the end of a spring compressed 2 inches. The linear spring

Question

A 5-lbm toy car is at the end of a spring compressed 2 inches. The linear spring has a force

constant of k = 3 lbf/in. (Recall that for a linear spring, F = k

A 5-lbm toy car is at the end of a spring compressed 2 inches. The linear spring has a force constant of k = 3 lbf/in. (Recall that for a linear spring, F = k middot x) Neglect friction in all parts. How much energy is stored in the compressed spring? Ans: 6lbf middot in If the spring is released, what is the maximum velocity of the car on a level surface in ft/s? Ans: V=2.54 ft/s If the car approaches a 15 degree incline, how much further will the car travel before it stops? Ans: s=0.386 ft If instead a motor is used to maintain a constant velocity on the incline, what horsepower is required? Ans: PE =0.00598 hp If instead the car approaches a 15 degree decline, what is the maximum velocity 5 feet further along the track? Ans: V = 9.43 ft/s

Explanation / Answer

m = 5/(9.8*39.37) = 0.012959lbin = 0.155517 lb f
a) W = 0.5*k*x^2 = 0.5*3*2^2 = 6lbf .in

b) 0.5*k*x^2 = 0.5*m*v^2

v = 2.54 ft/s

c) 0.5*k*x^2 = m g h

h = 6/5 = 1.2 in
sin15 = h/s

s = h/sin15 = 4.64 in

1 in = 0.08333 ft

s = 4.64*0.08333 = 0.386ft

d) w = 0.5*m*v^2
time = s/v = 0.386/2.54 = 0.152s
power = W/t = 6*0.08333/0.152 = 3.289 lbf/s

1hp = 550lb f /s

P = 0.00598 hp

e) h = s*sin15 = 1.29 ft

0.5*k*x^2 +mgh = 0.5*mv^2

6*0.08333+5*1.29 = 0.5*0.155517

v = 9.43 ft/s

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