(6%) Problem 4: A particular spring does not obey Hooke\'s law (F(x)--kx), but r
ID: 2041368 • Letter: #
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(6%) Problem 4: A particular spring does not obey Hooke's law (F(x)--kx), but rather is described by F(x) =-k1 x-k2x2 x is the displacement of the spring from equilibrium and the constants ki and k2 are given below Randomized Variables k1 74 N/m k2 = 7.2 N/m? 50% Part (a) Input an expression for the potential energy of the spring as a function of position. Assume that U(0)-0 Grade Summary 100% 7 8 9 HOME Submissions Attempts remaining: 8 (1% per attempt) detailed view 0 END BACKSPACE DECLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback. 50% Part (b) What is the magnitude of the restoring force of this spring, in newtons, if it is stretched 23.5 cm from equilibrium?Explanation / Answer
a)
we know, F = -dU(x)/dx
==> -dU(x) = F*dx
U(x) = -integral F*dx
= -integral (-k1*x - k2*x^2)*dx
= integral (k1*x + k2*x^2)*dx
= k1*(x^2/2) + k2*(x^3/3)
= k1*x^2/2 + k2*x^3/3 <<<<<<<<-----Answer
b) at x = 23.5 cm = 0.235 m
F(x) = -k1*x - k2*x^2
= -7.4*0.235 - 7.2*0.235^2
= -2.14 N
|F(x)| = 2.14 N <<<<<<<<-----Answer
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