Phosphorus bursts into flame when exposed to oxygen. What element might be used when storing pure phosphorus to keep it from rea
2 answers:
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Answer
given,
total mass = 18.3 Kg
inclined at = 19.8°
distance = 19.7 m
coefficient of friction = 0.5
a) Horizontal component of force forward,
Horizontal force backward = force of friction, f =μN.
Normal force, N = mg - F sinθ
Since the velocity is uniform, acceleration, a = 0
Net force, ΣF = 0
F cos θ+ -μ(mg -F sinθ) = 0
F (cosθ + μsin θ) = μmg
=80.76 N
(b) Work done by the rope on the sledge,
W = F.S cosθ = 80.76 x 19.7 x 0.9360
= 1497 J
(c) Mechanical energy lost due to friction,
E = W done against friction,
W' = f x S =F x S = 80.76 x 19.7 x0.9360
= 1497 J
we know velocity is uniform,.
W done by the applied force = W done against friction.
= 1440 J
Answer:
21.8°
Explanation:
Let's call θ the angle between BC and the horizontal.
Draw a free body diagram for each block.
There are 4 forces acting on block D:
Weight force P pulling down,
Normal force N₁ pushing perpendicular to AB,
Friction force N₁μ pushing parallel up AB,
and tension force T pushing parallel up AB.
There are 4 forces acting on block E:
Weight force P pulling down,
Normal force N₂ pushing perpendicular to BC,
Friction force N₂μ pushing parallel to BC,
and tension force T pulling parallel to BC.
Sum of forces on D in the perpendicular direction:
∑F = ma
N₁ − P sin θ = 0
N₁ = P sin θ
Sum of forces on D in the parallel direction:
∑F = ma
T + N₁μ − P cos θ = 0
T = P cos θ − N₁μ
T = P cos θ − P sin θ μ
T = P (cos θ − sin θ μ)
Sum of forces on E in the perpendicular direction:
∑F = ma
N₂ − P cos θ = 0
N₂ = P cos θ
Sum of forces on E in the parallel direction:
∑F = ma
N₂μ + P sin θ − T = 0
T = N₂μ + P sin θ
T = P cos θ μ + P sin θ
T = P (cos θ μ + sin θ)
Set equal:
P (cos θ − sin θ μ) = P (cos θ μ + sin θ)
cos θ − sin θ μ = cos θ μ + sin θ
1 − tan θ μ = μ + tan θ
1 − μ = tan θ μ + tan θ
1 − μ = tan θ (μ + 1)
tan θ = (1 − μ) / (1 + μ)
Plug in values:
tan θ = (1 − 0.4) / (1 + 0.4)
θ = 23.2°
∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.
