The rope and pulley have negligible mass, and the pulley is frictionle

Carole Yarbrough 2021-12-16 Answered

The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the 
8.00-kg block and the tabletop is μk=0.250. The blocks are released from rest. Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m.

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Expert Answer

Bertha Jordan
Answered 2021-12-17 Author has 37 answers
Given:
d=1.5m
μk=0.25
m1=8.0kg
m2=6.0kg
Required: v
For horizontal mass
According to work-energy theorem the velocity is given by
Wt=12m1v2(1)
But the net work is also given by
Wt=WtWf=Wtμkmgd(2)
Where : Wt is the work due to tension.
From (1) and (2) get that
Wtμkm1gd=12m1v2(3)
For vertical mass
According to work-energy theorem the velocity is given by
Wt=12m2v2(4)
But the net work is also given by
Wt=WwWt=m2gdWt(5)
From (4) and (5) get that
Wt+m2gd=12m2v2(6)
Adding (3) and (6) to get
m2gdμkm1gd=12(m1+m2)v2Rih>(7)
Substitution in (7) to get that
v2=6×9.8×1.58×9.8×1.5×0.257v=2.9ms

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Mollie Nash
Answered 2021-12-18 Author has 33 answers
Step1
Given that
mass m1=6 kg
mass m2=8 kg
coefficient of kinetic friction μk=0.250
displacement of mass 6 kg y=1.50m
Step2
Assume that the uniform acceleration in both masses is "a" and tension in the rope is "T".
Now for mass m1
effective downward force = 6g - T
m1a=6gT
1) 6a=6gT
Now for mass m2
friction force ( towards left) fk=μkm2g
fk=(0.25)(8)(g)
fk=2g
effective force on m2( towards right)=Tfk
m2a=Tfk
2) 8a=T2g
Adding equation (1) and (2) as:
(6a)+(8a)=(6gT)+(T2g)
14a=4g
a=4g14
a=2g7
Step3
Effective force on m1
F1=6a
F1=(6)(2g7)
F1=12g7
Work done on m, is
W=F1y
W=(12g7)(1.5)
W=18g7
Step4
Now by the use of work energy theorem
Work done = change in kinetic energy
W=12mv2
where v = speed
18g7=12(6)v2
v2=36g42
g= gravitational acceleration ( 9.8ms2)
v=(36)(9.8)42
v=2.989ms

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