A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch
1 answer:
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Answer:
τ = 132.773 lb/in² = 132.773 psi
Explanation:
b = 12 in
F = 60 lb
D = 3.90 in (outer diameter) ⇒ R = D/2 = 3.90 in/2 = 1.95 in
d = 3.65 in (inner diameter) ⇒ r = d/2 = 3.65 in/2 = 1.825 in
We can see the pic shown in order to understand the question.
Then we get
Mt = b*F*Sin 30°
⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in
Now we find ωt as follows
ωt = π*(R⁴ - r⁴)/(2R)
⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)
⇒ ωt = 2.7114 in³
then the principal stresses in the pipe at point A is
τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)
⇒ τ = 132.773 lb/in² = 132.773 psi
Answer: The correct option is Option b.
Explanation:
Power is defined as the rate of work done by an object.
Mathematically,
.....(1)
And work done is the product of force exerted on the object times the displacement covered by that object.
Mathematically,
Putting this value in above equation, we get:
where,
P = power = ?W
F = Force exerted = 10N
s = Displacement = 400cm = 4m (Conversion factor: 1m = 100 cm)
t = Time taken = 8s
Putting values in above equation, we get
Hence, the correct option is Option b.