8. A massless spring of constant 80 N/m is fixed on the left end of a level trac
ID: 1783690 • Letter: 8
Question
8. A massless spring of constant 80 N/m is fixed on the left end of a level track. A block of mass 0.7 kg is pressed against he spring and compresses the spring a distance 'd', in METERS as indicated in the figure. The block, initially at rest, is released and travel towards a circular loop-the-loop of radius 1.5 m. The entire track is frictionless except the 2.7 m between points 'A' and 'B'. The coefficient of friction between 'A' and 'B' is 0.34. Determine the minimum compressed distance 'd' such that the block arrives at point 'C' such that it's on the verge of falling. Hint: Normal force at 'C' if the block is on the verge of falling.
e | .3 VHL Central | Course (-0ivylearn.ivytech.edu/courses/737548/quizzes/1431558/take by Loan Quiz: Cpt 7 Homework × Syllabus for 30A-Fall 2017-F chegg study | Guided Solut + a popular amusement park nde has a radius of 2.5 m and is set into circular motion at an angular speed of -5 rad/s as shown in the figure. The floor drops away and the person remains suspended against the vertical wall. What is the minimum coefficient of friction necessary to keep the rider from nbox Help D Question 8 2 pts A massless spring of constant 80 N/m is fixed on the left end of a level track. A block of mass 0.7 ke is pressed against he spring and compresses the spring a distance 'd', in METERS as indicated in the figure. The block, initially at rest, is released and travel towards a circular loop-the-loop of radius 1.5 m. The entire track is frictionless except the 2.7 m between points 'A' and 'B. The coefficient of friction between 'A'and B' is 0.34. Determine the minimum compressed distance d' such that the block arrives at point 'C' such that it's on the verge of falling. Hint: Normal force at C' if the block is on the verge of falling. 1:10 PM O Type here to search 10/19/2017Explanation / Answer
7)
Find the centripetal force on the rider from the wall (the "normal force"):
N = mv^2 / r = m(r)^2 / r = mr^2
Now, in order not to fall, the frictional force on the rider from the wall must equal the rider's weight:
Ff = N = mg
Now solve for :
= mg / N = mg / (mr^2) = g / (r^2)
= (9.8) / ((2.5)(5)^2) = 0.1568
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