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A small block of mass 5kg placed on the block of 20 kg and length is 25cm.block of mass 5 kg is placed at the right end of 20kg block.if a constant force of 15n is applied on block of 20kg horizanally the surfaces assumed 2 b frictionless.find time after whick block of mass 5kg sepreates frm 20 kg block
Dear jauneet surface is friction less so no horizontal force will act on the 5 kg block. it will remain at rest. acceleration of the 20 kg block = F/m = 15/20 = 3/4 time taken by 20 kg block to travel 25 cm distance is S= 1/2 at2 t= (2S/a)1/2 = (2*.25/3/4)1/2 = (2/3)1/2 sec Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
Dear jauneet
surface is friction less so no horizontal force will act on the 5 kg block. it will remain at rest.
acceleration of the 20 kg block = F/m = 15/20 = 3/4
time taken by 20 kg block to travel 25 cm distance is
S= 1/2 at2
t= (2S/a)1/2
= (2*.25/3/4)1/2
= (2/3)1/2 sec
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
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