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gravel is dropped into a conveyer belt at rate of 0.5kg/sec. the xtra force needed to keep the belt moving at 2m/s is?
Dear Kedar The net force should be defined as the rate of change of momentum here v is constant hence dv/dt is 0 dm/dt= 0.5 kg/sec, v=2m/s hence extra force required= v*dm/dt = 2*0.5 = 1N All the best. AKASH GOYAL AskiitiansExpert-IITD Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
Dear Kedar
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
This is the case of same velocity and changing mass. So F=-u*(dm/dt)= 2*0.5=1N <F-dp/dt=d/dt(mv)=v(dm/dt)+m(dv/dt)= v*(dm/dt) bcoz vel remains constt, dv/dt=0> So a force equal to newton is required. DO APPROVE I YOU LIKE IT DEEKSHA
This is the case of same velocity and changing mass.
So F=-u*(dm/dt)= 2*0.5=1N <F-dp/dt=d/dt(mv)=v(dm/dt)+m(dv/dt)= v*(dm/dt) bcoz vel remains constt, dv/dt=0>
So a force equal to newton is required.
DO APPROVE I YOU LIKE IT
DEEKSHA
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