A force is applied to an ideal spring (initially in its equilibrium position) and does 1.9 JJ of work stretching it 2.2 cmcm . H
2 answers:
Answer:
W=16.58J
Explanation:
initial information we have
work:
stretched distance:
from this, we can find the value of the constant of the spring k, with the equation for work in a spring:
substituting known values:
and clearing for k:
and now we want to know how much work is done when we stretch the spring a distance of 6.5cm from equilibrium, so now x is:
and using the same formula for work, with the value of k that we just found:
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Answer:
Explanation:
Work: This can be defined as the product of force and distance. The unit of work is Joules (J). it can be expressed mathematically as
W = F×d
or
W = .................................. Equation 1
Where b = upper limit, a = lower limit, Fx = expression of force.
<em>Given: a = 0 , b = 1.3 m, Fx = 4 + 15.7x - 1.5x²</em>
Substituting these values into equation 1
<em>W = </em>
W = ᵇ[4x + 15.7x²/2-1.5x³/3 +C]ₐ
Work = upper limit - lower limit
Work = ᵃ[4x + 15.7x²/2 - 1.5x³/3 +C] - [4x + 15.7x²/2 + 1.5x³/3 +C]ᵇ............... Equation 2
Substituting the values of a and b into equation 2
Work = [4(1.3) + 15.7(1.3)²/2-1.5(1.3)³/3 + C] - [0 +C]
Work = [5.2 + 26.53 -3.29 + C] - C
Work = 28.44 J
Work done by the force = 28.44 J.