Question Part Points Submissions Used 1 2 3 –/1 –/1 –/1 0/3 0/3 0/3 Total –/3 .T

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Question Part
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Submissions Used
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0/3 0/3 0/3
Total
–/3

.This question has some parts which are scored using special conditions.
Question Parts 1, 2, 3: A bonus of 10% of the points for this question will be added for each question submitted more than 9 hours before the due date.
..Objects of masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley as in the figure below. The object m1 is held at rest on the floor, and m2 rests on a fixed incline of ? = 45.0°. The objects are released from rest, and m2 slides 1.60 m down the slope of the incline in 4.70 s.


Determine the acceleration of each object

Determine the tension in the string
Determine the coefficient of kinetic friction between m2 and the incline

Explanation / Answer

Given that distance d=1.60m time taken t=4.70s using kinematic equation equation : d = ut+(1/2)at^2 acceleration a can be determined
1.60 = 0 + 0.5a(4.702)
1.60 = 11.04a
a = 0.1449 m/s2

consider m1 :      F = m1a
-m1g+T = m1a
         T = m1g+m1a             = m1(g+a)
T = 4(9.8+0.1449)
T = 39.779 N
for m2:
m2gsin - T - f = m2a, so                    f =m2gsin - T - m2a
N = m2gcos
= f/N       = (m2gsin - T -m2a)/(m2gcos)       = 21.17/62.36       =0.339
please rate if any mmistake in this problem please send mail I will clarify your doubt.
Given that distance d=1.60m time taken t=4.70s using kinematic equation equation : d = ut+(1/2)at^2 acceleration a can be determined
1.60 = 0 + 0.5a(4.702)
1.60 = 11.04a
a = 0.1449 m/s2

consider m1 :      F = m1a
-m1g+T = m1a
         T = m1g+m1a             = m1(g+a)
T = 4(9.8+0.1449)
T = 39.779 N
for m2:
m2gsin - T - f = m2a, so                    f =m2gsin - T - m2a
N = m2gcos
= f/N       = (m2gsin - T -m2a)/(m2gcos)       = 21.17/62.36       =0.339
please rate if any mmistake in this problem please send mail I will clarify your doubt.

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