Please wait...
Contact Us
Contact
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

93219 24448 / 99871 78554

Mon to Sat - 10 AM to 7 PM

sir please answer

qsnImg
Asked by minipkda 24th May 2018, 6:57 AM
Answered by Expert
Answer:
 
Figure shows the force acting on the block which is at an inclined plane. Force F shown in figure is due to the applied force of constant power. 
 
Block moves with constant velocity without any acceleration. hence Forces are balanced.
 
hence we have F = mgsinθ +μN = mgsinθ + μmgcosθ 
 
required fraction is given by
 
begin mathsize 12px style fraction numerator m g sin theta over denominator m g sin theta plus 0.25 cross times m g cos theta end fraction space equals space fraction numerator sin begin display style theta end style over denominator sin begin display style theta end style begin display style plus end style begin display style 0 end style begin display style. end style begin display style 25 end style begin display style cos end style begin display style theta end style end fraction
end style
θ = 37º (given). After substituting for θ in the above equation we get the fraction 0.75 
Answered by Expert 25th May 2018, 11:21 AM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Tags: power
Your answer has been posted successfully!

Chat with us on WhatsApp