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An ideal massless spring S can be compressed 1m by a force of 100N in equilibrium. The same spring is placed at the bottom of a frictionless inclined plane inclined At 30° to the horizontal. A 10 kg block M is released from the top of incline and is brought to rest momentarily after compressng the spring by 2m.if g=10 m/s2.what is the speed of mass just before it touches the spring
It is given that force of 100N is applied for compression of 1m.F=kx100=k(1) So, k=100N/mNow, the whole kinetic energy of the block is converted into spring potential energy.So,1/2mv^2=1/2kx^2Cancelling 1/2 from both the sides.mv^2=kx^2then 10*v^2=100(2)^2 . Becuase it is given that it compresses the spring by 2mso,v^2=100*4/10v^2=40so, v = root 40 which is the answer HOPE IT CLEARS ALL THE BEST
Sir there we want to take the consideration of potential energy also I think..the PE and KE of block is converted into spring PE???plz explain once more
No we don`t need to consider the PE of the block because given that the spring is at the bottom of the inclined plane, it means that we need to consider the height to be zero so PE is zero
But in the solution of this question PE is also taken into consideration and the answer is root40. Its a question from objective chemistry for neet
I think the answer I root 40.You should consider kinetic energy just before coming in contact with spring and potential energy after compression and equate them( law of conservation of energy).
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