A small mass slides down an inclined plane of inclination θ = 6 0 ∘ with respect to the horizontal. The acceleration of the block varies as a = k x 2 , where x is the distance through which the mass slides down and k is a constant. The coefficient of friction is 1 when the block covers a distance of 1 0 . Find the value of k .
Details and Assumptions
:
Take
g
=
1
0
m/s
2
.
Every quantity is in SI unit.
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Same method, easy problem.
Same way buddy
My ans is coming as 3times of your ans I have used work energy theorem pls check
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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
My solution is correct.. Show me your work first to check it :)
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.
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u is not constant!!!
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Read properly it's given 1
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@Rahul Malhotra – Its 1 when it covers the distance not always.I said rightly.
a = m F = m 5 ( 3 − 1 ) m = k ( 1 0 ) 2
⟹ k = 1 0 5 ( 3 − 1 ) ≈ 0 . 3 6 6
Answer: 0 . 3 6 6
Caution: a = 0 . 3 6 6 x 2 does not reflect a logical physics for the whole journey.
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photo To know direction of forces... I solved it by Newton's second law ∑ F = m a = 1 0 m sin 6 0 − 1 0 μ m cos 6 0
Now we have a = 5 + 5 3 this value we have when μ = 1 This mean the distance is 1 0 Finally we have 5 + 5 3 = k × ( 1 0 ) 2 so k = 2 3 − 1 ≈ 0 . 3 6 6