(3%) Problem 17: A block of mass m1-2.5 kg is attached to a spring with spring c
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(3%) Problem 17: A block of mass m1-2.5 kg is attached to a spring with spring constant k = 510 N/m. It is initially at rest on an inclined plane that is at an angle of = 270 with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is ,-0.15. In the initial position, where the spring is compressed by a distance of d= 0.15 m, the mass is at its lowest position and the spring is com the block as zero pressed the maximum amount. Take the initial gravitational energy of Randomized Variables m=2.5 kg k = 510 N/m d=0.15 m =270 Uz=0.15 ©theexpertta.com 50% Part (a) What is the block's initial mechanical energy, E 0 in J? * 50% Part (b) If the spring pushes the block up the incline, what distance, L in meters, will the block travel before coming to rest? The spring remains attached to both the block and the fixed wall throughout its motion Grade Summary Deductions Potential 2% 98% Submissions Attempts remaining: 6 (296 per attempt) cos) cotan)asin)acos) atan()acotan)sinhO cosh0 tanh0 cotanh0 sin tan 4 5 6 detailed view 2% 0 END Degrees Radians DEL CLEAR Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 9 Feedback: i for a 0% deduction At the point L the system not only has potential energy of gravity, but also potential elastic energy. You may have forgotten the term of elastic energy at the final point. Submission History Answer Hints Feedback Totals 2% 0% 0% At the point L the system not only has potential energy of gravity, but also potential elastic energy. You may have forgotten the term of elastic energy at the final point. 2% 0.17 Totals 2% 0% 0% 2%Explanation / Answer
B)
By work energy theorem,
Wncon = delta(PEg) + delta(KE) + delta(PEs)
where PEg = gravitational potential energy, and PEs = potential energy due to the spring
Thus,
-uk M g L = -M g L sin 27 + 1/2 k (L^2 - d^2)
-0.15 * 2.5 * 9.8 * L = -2.5 * 9.8 * L* sin 27 + 0.5 * 510 * (L^2 - 0.15^2)
- 3.675 * L = - 11.12 * L + 255 * (L^2 -0.15^2)
- 3.675 * L = - 11.12 * L + 255 L^2 - 5.74
255 L^2 - 7.445 L - 5.74 = 0
solving for L,
L = 0. 165 m
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