Determine the velocity of the 60-lb block A if the two blocks are released from rest and the 40-lb block B moves 2 ft up the inc
line. The coefficient of kinetic friction between both blocks and the inclined planes is μk=0.10.
1 answer:
vₐ = 0.771 ft/s
vb = -1.54 ft/s
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<u>Explanation:</u>
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Block A:
F = ma
Nₐ = - 60 cos60° = 0
Nₐ = 30 lb
Fₐ = 0.1 X 30 = 3lb
Block B:
F = ma
Nb = - 40 cos30° = 0
Nb = 34.64lb
Fb = 0.1 X 34.64 = 3.464lb
T1 + ∑U = T2
(0 + 0) + 60 sin 60° |Δsₐ| - 40 sin 30° |Δsb| - 3|Δsₐ| - 3.464 |Δsb|
= 1/2 (60/32.2) vₐ² + 1/2 (40/32.2) vb²
and
2vₐ = - vb
On solving, we get
vₐ = 0.771 ft/s
vb = -1.54 ft/s
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Answer:
a) the coefficient of performance of the air conditioner is 3.5729
b)
- the power input required for a reversible air conditioner is 0.645 hp
- the coefficient of performance for the reversible air conditioner is 18.2759
Explanation:
Given the data in the question;
Lower Temperature = 70°F = ( 70 + 460 )R = 530 R
Higher Temperature = 99° F = ( 99 + 460 )R = 559 R
Cooling Load = 30000 Btu/h
we know that 1 hp = 2544.43 Btu/h
Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h
a)
Coefficient of performance of the air conditioner;
= Cooling Load
/ power P
we substitute
= 30000 Btu/h / 8396.619 Btu/h
= 3.5729
Therefore, the coefficient of performance of the air conditioner is 3.5729
b)
- Power input required ( in hp )
/
=
/ (
-
)
we substitute
30000 Btu/h / = 530 R / ( 559 R - 530 R )
30000 Btu/h / = 530 R / 29 R
we solve for
= ( 30000 Btu/h × 29 R ) / 530 R
= ( 870000 Btu/h / 530 )
= 1641.5094 Btu/h
we know that; 1 hp = 2544.43 Btu/h
so;
= ( 1641.5094 / 2544.43 ) hp
= 0.645 hp
Hence, the power input required for a reversible air conditioner is 0.645 hp
- the coefficient of performance for the reversible air conditioner;
=
/ (
-
)
we substitute
= 530 R / ( 559 R - 530 R )
= 530 R / 29 R
= 18.2759
Hence, the coefficient of performance for the reversible air conditioner is 18.2759