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A block of mass m is at rest at the origin at t = 0. It is
Chapter 6, Problem 72CP(choose chapter or problem)
A block of mass \(m\) is at rest at the origin at \(t=0 .\) It is pushed with constant force \(F_{0}\) from \(x=0\) to \(x=L\) across a horizontal surface whose coefficient of kinetic friction is \(\mu_{\mathrm{k}}=\mu_{0}(1-x / L)\). That is, the coefficient of friction decreases from \(\mu_{0}\) at \(x=0\) to zero at \(x=L\).
a. Use what you've learned in calculus to prove that
\(a_{x}=v_{x} \frac{d v_{x}}{d x}\)
b. Find an expression for the block's speed as it reaches position \(L\).
Equation Transcription:
Text Transcription:
m
t = 0
F_0
x = 0
x = L
mu_k = mu_{0}(1-x / L)
mu_{0}t
x = 0
x = L
a_{x} =v_{x} frac{d v_{x}}{d x}
L
Questions & Answers
QUESTION:
A block of mass \(m\) is at rest at the origin at \(t=0 .\) It is pushed with constant force \(F_{0}\) from \(x=0\) to \(x=L\) across a horizontal surface whose coefficient of kinetic friction is \(\mu_{\mathrm{k}}=\mu_{0}(1-x / L)\). That is, the coefficient of friction decreases from \(\mu_{0}\) at \(x=0\) to zero at \(x=L\).
a. Use what you've learned in calculus to prove that
\(a_{x}=v_{x} \frac{d v_{x}}{d x}\)
b. Find an expression for the block's speed as it reaches position \(L\).
Equation Transcription:
Text Transcription:
m
t = 0
F_0
x = 0
x = L
mu_k = mu_{0}(1-x / L)
mu_{0}t
x = 0
x = L
a_{x} =v_{x} frac{d v_{x}}{d x}
L
ANSWER:
Step 1 of 4
We need to prove the given relation using chain rule
We are required to derive an expression for the block’s speed as it reaches position L