Q: A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What is its stopping distance on a roadway sloping downward at an angle of 19.0°?
If I calculated correctly, I found that u_k = 0.9776 on a flat plane. For the inclined plane I found that
(Sum)F_y = N' - mgsin(theta) and
(Sum)F_x = f_k' - mgcos(theta)
How do I proceed, I am so confused!! Please help!! Thanks!
If I calculated correctly, I found that u_k = 0.9776 on a flat plane. For the inclined plane I found that
(Sum)F_y = N' - mgsin(theta) and
(Sum)F_x = f_k' - mgcos(theta)
How do I proceed, I am so confused!! Please help!! Thanks!
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flat surface:
F = m a = u mg so u = a/g
vf^2=v0^2 + 2ad
vf=0
v0=60mi/hr = 26.9m/s
a=accel
d=distance = 123ft = 37.5m
0=26.9^2 + 2*a*37.5 => a = - 9.65m/s/s
u = a/g = 0.98
on the incline:
sum of forces = ma
forces are component of gravity down the plane = mg sin(theta) and
friction up the plane = u mg cos(theta)
therefore, newton's second law becomes
mg sin(theta) - u mg cos(theta) = ma
a = 9.8(sin 19 - 0.98 cos 19) = -5.9m/s/s
and now use
vf^2=v0^2 + 2ad
vf=0
v0=26.9m/s
a=-5.9m/s/s
solve for d:
0=26.9^2 - 2*(5.9m/s/s)*d => d= 61.4m
F = m a = u mg so u = a/g
vf^2=v0^2 + 2ad
vf=0
v0=60mi/hr = 26.9m/s
a=accel
d=distance = 123ft = 37.5m
0=26.9^2 + 2*a*37.5 => a = - 9.65m/s/s
u = a/g = 0.98
on the incline:
sum of forces = ma
forces are component of gravity down the plane = mg sin(theta) and
friction up the plane = u mg cos(theta)
therefore, newton's second law becomes
mg sin(theta) - u mg cos(theta) = ma
a = 9.8(sin 19 - 0.98 cos 19) = -5.9m/s/s
and now use
vf^2=v0^2 + 2ad
vf=0
v0=26.9m/s
a=-5.9m/s/s
solve for d:
0=26.9^2 - 2*(5.9m/s/s)*d => d= 61.4m