Question

O Quit: Honors Work, Power, and「 × O Farris Rafique'sQuiz History. Hox | Σ work calculator isdinstructure.com/courses/162718/quizzes/144982/take D | Question 29 0.5 pts Hi 3.13 A 9.7 kg box is held against a spring which is compressed 1.2 m as shown on top of an incline which is 4.1 m (Hg). Find the minimum spring constant k (N/m) so that the box just makes it to the top of the 10.8 m (H2) high second incline. All surfaces are frictionless except the crosshatched surface where d-18.1 m and μί-0.1 and the incline where ,n-os 5-9814 HINT: Use the Conservation of Energy Equation ME ME Think of the initial ME as gravitational potential energy plus elastic potential energy And then subtract the work lost due to friction on the flat surface. And then subtract the work lost due to friction on the incline plane. Next. set this initial ME equal to final ME which is all gravitational potential at the stopping point Finally, solve for k N2

Answer 1

Solve this problem using conservation of energy.

Spring constant of the spring = k N/m

Initial compression, x = 1.2 m

Total initial energy of the block, TE1 = spring energy + Potential energy of the block

= (1/2)*k*x^2 + m*g*H1

Now, consider the final condition -

Work done against the friction, Wf1 = $\mu 1$*m*g*d = 0.1*9.7*9.81*18.1 = 172.2 J

Potential energy achieved by the block = m*g*H2 = 9.7*9.81*10.8 = 1027.7 J

Length of the incline, L = H2 / sin53.13 = 10.8/sin53.13 = 13.5 m

Work done against the friction on the inclined plane, Wf2 = $\mu 2$*m*g*cos53.13*L = 0.5*9.7*9.81*cos53.13 * 13.5

= 385.4 J

Apply conservation of energy -

(1/2)*k*x^2 + m*g*H1 = 172.2 J + 1027.7 J + 385.4 J

=> 0.5*k*1.2^2 + 9.7*9.81*4.1 = 1585.3

=> 0.72*k = 1195.2

=> k = 1195.2 / 0.72 = 1660 N/m (Answer)