InterviewSolution
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InterviewSolution
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An ideal massless spring `S` can compressed `1.0` m in equilibrium by a force of `1000 N`. This same spring is placed at the bottom of a friction less inclined plane which makes an angle `theta =30^@` with the horizontal. A (10 kg) mass (m) is released from rest at the top of the incline and and is brought to rest momentarily after compressing the spring by `2 m. if `g =10 ms^(-1)`, what is the speed of just before it touches the spring? .A. (a) `sqrt20 ms^(-1)`B. (b) `sqrt30 ms^(-1)`C. (c) `sqrt10 ms^(-1)`D. (d) `sqrt40 ms^(-1)` |
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Answer» Correct Answer - A `F=kx` `:. k=(F)/(x) =(100)/(1)` `=100 N//m` `E_(i)=E_(f)` `:. 1/2 xx 10 xx v^(2) =1/2 xx 100 xx (2)^(2)-(10)(10)(2 sin 30^@)` Solving we get, `v=sqrt20 m//s` |
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