The body slides off an inclined plane

Thread starter Adeopapposaurus
Start date Apr 18, 2021
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Body Incline plane Inclined Inclined plane Plane Slide

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The discussion focuses on the equations governing the motion of a body sliding off an inclined plane, specifically addressing the forces involved, including gravitational force and friction. The key equations presented are for acceleration and velocity, with a calculated friction coefficient of 0.26. A user seeks verification of their solution, which appears to be correct based on the responses received. The calculations involve using specific values for gravitational acceleration and angles. Overall, the thread confirms the accuracy of the user's approach to solving the problem.

Apr 18, 2021

Adeopapposaurus

Homework Statement: The body slides off an inclined plane with an inclination angle of α = 60°. After t = 2s, the body speed is v = 14.7 m/s. Find the friction coefficient f. The initial velocity of the body is 0 m/s and the gravity acceleration g = 10 m/s^2 .

Relevant Equations: F = f * N
F = fmg*cos(α)
ma = mg * sinα - fmg * cosα

ma = mg * sinα - fmg * cosα
a = g (sinα-f * cosα)
v = g*t(sinα - f * cosα)
14.7 = 10 * 2 (sin60 - f * cos60)
f = 0.26

Can someone please check if my solution is correct? I'd really appreciate that.

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Apr 18, 2021

PeroK

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Looks right.

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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