Sliding Down An Inclined Plane

A small mass slides down an inclined plane of inclination θ = 6 0 \theta = 60^\circ with respect to the horizontal. The acceleration of the block varies as a = k x 2 a = k x^{2} , where x x is the distance through which the mass slides down and k k is a constant. The coefficient of friction is 1 when the block covers a distance of 10 \sqrt{10} . Find the value of k k .

Details and Assumptions :
Take g = 10 m/s 2 g=10 \text{ m/s}^2 .
Every quantity is in SI unit.


The answer is 0.366.

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2 solutions

Refaat M. Sayed
Feb 16, 2016

photo To know direction of forces... I solved it by Newton's second law F = m a = 10 m sin 60 10 μ m cos 60 \displaystyle\sum F = ma = 10m \sin 60 - 10\mu m \cos 60

Now we have a = 5 + 5 3 a= 5+5\sqrt {3} this value we have when μ = 1 \mu=1 This mean the distance is 10 \sqrt {10} Finally we have 5 + 5 3 = k × ( 10 ) 2 5+5\sqrt {3}=k \times (\sqrt{10})^2 so k = 3 1 2 0.366 k = \frac{\sqrt{3 } -1}{2} \boxed {\approx 0. 366}

Same method, easy problem.

Adarsh Kumar - 5 years, 3 months ago

Same way buddy

Kaustubh Miglani - 5 years, 3 months ago

My ans is coming as 3times of your ans I have used work energy theorem pls check

Rahul Malhotra - 5 years, 3 months ago

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By integrating acc. v^2=[20 [10]^0.5]/3 *K Putting the value of v^2 in the eq ghsin60-gcos60 [10]^0.5=0.5[v]^2

Rahul Malhotra - 5 years, 3 months ago

My solution is correct.. Show me your work first to check it :)

Refaat M. Sayed - 5 years, 3 months ago

This question is incorrect as if we use nlm eq Mgsin60-umgcos60=ma Remove m from the equation and the value comes out to be a constant. This means that acc is not a function of "x ". Hope u understand.

Rahul Malhotra - 5 years, 3 months ago

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u is not constant!!!

Kaustubh Miglani - 5 years, 3 months ago

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Read properly it's given 1

Rahul Malhotra - 5 years, 3 months ago

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@Rahul Malhotra Its 1 when it covers the distance not always.I said rightly.

Kaustubh Miglani - 5 years, 3 months ago
Lu Chee Ket
Feb 17, 2016

a = F m = 5 ( 3 1 ) m m = k ( 10 ) 2 a = \frac{F}{m} = \frac{5(\sqrt3 -1) m}{m} = k (\sqrt{10})^2

k = 5 ( 3 1 ) 10 0.366 \implies k = \frac{5(\sqrt3 - 1)}{10} \approx 0.366

Answer: 0.366 \boxed{0.366}

Caution: a = 0.366 x 2 a = 0.366~x^2 does not reflect a logical physics for the whole journey.

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