Question

Only one try left. Thought I knew it. Please use 5.8 lb.

The 5.8-lb cylinder is falling from A with a speed vA 10 ft/s onto the platform. The spring has an unstretched length of 1.75 ft and is originally kept in compression by the 1-ft long cables attached to the platform. Neglect the mass of the platform and spring and any energy lost during the collision. (Figure 1) ▼ Part A Determine the maximum displacement of the platform, caused by the collision. Express your answer to three significant figures and include the appropriate units. 0.074 ft Submit Incorrect; One attempt remaining: Try Again Provide Feedback Figure 1 of 1

Answer 1

Calculate the work done on spring force of the block. ks Here, s, and s, are the position of the varied spring, k is the spring constant. Substitute 400lb/ft.for the value of k. -(*) -ks, 2 sp G(100)(1.75 –1) + s)' –(400)(1.75 –1) U, =-(400)(0.75 + s) -;(400)(0.75)| U, =-(200(0.75+s))+112.5 Let s be the distance moved by the spring after the 5.81bcylinder falls from a height of 3 m. Calculate the work done by the spring force. = Fx U, = 5.8(3+ s) Calculate the maximum displacement of the platform caused by the collision using the principle of work and energy equation. T, +EU_2 = T, +U,)=0 sp 2 Here, v is the speed of the cylinder falling from point A. Substitute 5.8(3 + s) for U -(200(0.75 + s)*) +112.5 forU, 5.8 -slugs for m and 10ft/s for v. 32.2 mv² +(Uz, +U;)=0 «(10)+ (s8(3+a)+[-(200(0.75 + s))+112.5])| 5.8 32.2X( 9+17.4 +5s +(-112.5 – 200s? –300s +112.5) = 0 24 +5s – 112.5 – 200s – 300s+112.5 =0 200s? +295s – 26.4 =0 s = 0.0846 ft