This is the 9th time the question was answered wrong by the ordinary physics exp
ID: 1343632 • Letter: T
Question
This is the 9th time the question was answered wrong by the ordinary physics expert. The answer was 6.9 m/s squared , but they kept writing it in a wrong situation. Now this time, please have an advanced expert in physics solve this problem correctly with the right situation according to the practice exam question 3 problem based on the link below. thank you. Please call me if interested for further assistance to helping me solve problem if applicable.
Hint: the free body diagram is seperate for each person relevant to the situation. for paul it is -T + Mpg=msa. Please copy and paste the link below and scroll down to the practice question 3 to get a clear picture over what I mean.
https://fiu.blackboard.com/bbcswebdav/pid-3936889-dt-content-rid-37965495_1/courses/1158-FIU01-PHY-2048-SECU01-82467/1158-FIU01-PHY-2048-SECU01-82467_ImportedContent_20150820015334/PHY2048_Exam_2_F2014.pdf
Paul ( mass mp=80kg ) accidently falls off the edge of a glacier .Fortunately, Paul is tied by a long, mass-less rope to steve ( mass ms= 80kg ), who has a climbing ax. Before Steve sets his ax to stop them, he accelerates down the ice with paul attached to the rope. Assume that the angle is 30 degrees, the kinetic coefficient of friction between steve's clothes and the ice is 0.10, and the friction between the rope and ice is negligible.
a. Use newton's 2nd law to write the equations of motion for each person. To recieve credit, you must draw a free body diagram for each person. Be sure to include a coordinate system and clearly label all forces for each diagram.
b. Determine the acceleration of each person.
Please show your work clearly and write step by step solution. Include also numeric substitution and variable manipulation
Explanation / Answer
use the summ of forces as
net force F = N = m*g*cos 30 for both
f= miu*N
---------------------------
b)
accelration a = F/m
a = Net force along the incline / total mass
a =(m1*g*sin 30 + m2*g*sin 30 -f1-f2) / (m1+m2)
f1 = miu*m1*g*cos 30
f1 = 0.1*80*9.8*cos 30
f1 = 67.9 N
f2 = miu*m2*g*cos 30
= 0.1*80*9.8*cos 30
f2 = 67.9 N
a = Net force along the incline / total mass
=(m1*g*sin 30 + m2*g*sin 30 -f1-f2) / (m1+m2)
= (80*9.8*sin 30+80*9.8*sin 30 - 67.9-67.9)/(80+ 80)
= 4.02 m/s^
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