Question

A 5-kg collar A is at rest on top of, but not attached
to, a spring with stiffness k1 = 400 N/m when a constant
150-N force is applied to the cable. Knowing A has a speed
of 1 m/s when the upper spring is compressed 75 mm,
determine the spring stiffness k2. Ignore friction and the
mass of the pulley

Answer 1

Apply the principle of work and energy to the collar A. Work and energy principle states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement Hence, according to the work and energy principle. T, +U,-2 =T, Here, the initial kinetic energy of the collar is T. the final kinetic energy of the collar A when the upper spring is compressed to 75 MM is T. and the total work done on the collar is U-2:

Initially the collar A is at rest, so the initial kinetic energy of the collar (T) is zero. Hence, T, = 0. Express the kinetic energy of the collar A when the upper spring is compressed to 75 MM (72). Here, the mass of the collar is m,, and velocity of the collar is va Substitute 5 kg for me. and 1 m/s for va T: = (5 kg)(1 m/s) = 2.5 J

Step 3 of 7 A Calculate the work done by the collar A as it is raised from level A to level B [(4,-2).]. (0-2), =-migh Here, the distance between the upper and lower spring is h. and acceleration due to gravity is 8 Substitute 5 kg for m,. 9.81 m/s? for g. and 0.45 m for h. (0.-2), - - (5 kg)(9.81 m/s*)(0.45 m) = -22.0725 J Force applied by the spring i at initial position is equal to the weight of the collar. F, = m,8 Here, the force applied by the spring I at initial position is Fi. Substitute 5 kg for me. and 9.81 m/sfor g. F = -(5 kg)(9.81 m/s) = -49.05 N Express the spring force when the collar moves up through a distance of x,. F = F, -k, x, Here, k, is the spring coefficient, and the force applied by the spring is F.

Calculate the distance at which the collar separates from spring' (x,). E *, ki Substitute 49.05 N for F, and 400 N/m for k.. 49.05 * 400 = 0.122625 m Calculate the work force exerted by the spring I as the collar separates from the spring [(4-2)] (U-2), =S" (F, – k,x ]dx =Fır, - 3*,*} -kv-{ Substitute 400 N/m for k, and 0.122625 m for x,. (0,1-2), = (400 N/m)(0.122625 m) = 3.0074 )

Express the work force exerted by the spring when it is compressed to 0.075 m [(4,-2)]. (U1-2); -- Ky Here, k, is the spring coefficient of the spring 2 and y is the distance of the compressed upper spring Substitute 0.075 M for y. (U-2); ----, (0.075 m) =-0.0028125k

Calculate the work of the 150 N cable force (U.-2), (U-), =(150 N): ..... (2) Here, the displacement of the 150 N force is d. In the initial position, when the collar is resting on spring 1. calculate the length of AB by using Pythagorean theorem. ('.x), = J(450 mm)' +(400 mm) 362,500 = 602.08 mm In the second position, when the upper spring is compressed, the length AB is 400 MM. ('.), = 400 mm Calculate the displacement of the 150 N force (d). d = ('8), -('.)2 Substitute 602.8 MM for (118), and 400 mm for ('.n), d = 602.08 mm - 400 mm = 202.08 mm Substitute 0.20208 m for din Equation (2). (U,-2), = (150 N)(0.20208 m) = 30.312 J

Calculate the total work done (U.-2). (U-2) = (U,-z), +(U-2), +(U-2), +(U.-2), Substitute -22.0725 for (U.-2),.3.0074 J for (U1-2);: -0.0028125k, J for (U-2)ę. and 30.3123 for (U-2)p (U-2)=-22.07251 +3.0074 J -0.0028125k, +30.312 J = 11.247 -0.0028125k, Substitute O for T,. 11.247 -0.0028125k, for U.-2. und 2.5 J for T, in Equation (1). T, +U -2 = T, 0+11.247 -0.0028125k, = 2.5J 0.0028125k, = 11.247 - 2.5 0.0028125k, = 8.474 8.474 ky 0.0028125 = 3110 N/m Hence, the stiffness of the spring 2 is 3110 N/m.

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