(7.2) A spring is positioned on a frictionless ramp, as shown. A force of 260 N
ID: 1785968 • Letter: #
Question
(7.2) A spring is positioned on a frictionless ramp, as shown. A force of 260 N is required to compress the spring by 0.020 m. A 12.5-kg box is released from rest on the incline and, after interacting with the spring, compresses the spring by 0.045 m when it first comes to rest. F, (N) 260 ---- (a) Calculate the spring constant. (b) Determine how far down the incline the box travels from its starting point to the location of maximum compression. (c) Based on your answer to part (a), what is the speed of the box at the instant it first touches the spring. Use conservation of energy to solve this exercise. Answers: (a) 13 x 103 N/, (b) 0.215 m, (c) 1.29 m/sExplanation / Answer
(a) Use the expression -
F = K*x
where K is the spring constant of the spring.
So, 260 = K*0.020
=> K = 260/0.020 = 13000 N/m = 13 x 10^3 N/m
(b) First determine the potential energy lost by the box and from this, deterime the height descended by the box.
(1/2)K*x^2 = mgh
=> 0.5*13000*0.045^2 = 12.5*9.81*h
=> h = 0.1073 m = 10.73 cm.
So, the distance along the inclined surface the box travelled, l = h / sin30 = 10.73/0.5 = 21.46 cm = 21.5 cm
= 0.215 m
(c) (1/2)K*x^2 = (1/2)m*v^2
=> 0.5*13000*0.045^2 = 0.5*12.5*v^2
=> v^2 = (13000*0.045^2) / 12.5
=> v = 1.45 m/s
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