I have a question based off of the response to the question I posted below. Sorry, i wanted to make sure that i clearly understood,so My question about the third solution is: Is the horizontal force zero N? or is it 721.3N?
A dragster and driver together have mass 918.8 kg. The dragster,starting from rest,attains a speed of 25.8 m/s in .53s .
Find the average acceleration of the dragsters during this time interval. Answer in units of m/s^2.
What is the size of the average force on the dragster during this time interval?
Assume the driver has a mass of 73.6kg. What horizontal force does the seat exert on the driver?
Here is my work,but i don't know if i did it correctly:
Vf=Vi+a(delta t)
25.8=0+a(.53)
48.68=a
918.8(48.68)=44,726.5=F
73.6(48.68)=3,582.8N
physics - Henry, Sunday, July 3, 2011 at 6:14pm
a = 25.8 / 53 = 0.487m/s
F = 918.8*0.487 = 447.5N.
Fd = mg = 73.6kg * 9.8N/kg = 721.3N.
The angle between seat and hor = 0 deg.
Fh = 721.3sin(0) = 0 Newtons.
Fv = 721.3cos(0) = 721.3N = ver. force of driver.
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