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Answer:
the unloaded braking efficiency is 84.6 %
Explanation:
Given the data in the question;
by Ignoring aerodynamic resistance; we can find the theoretical stopping distance using the following formula
S = (Y( V₁² - V₂²)) / ( 2g( ηbμ +
± sin∅
))
now given that the tracked is levelled, ∅ = 0, also Y
= 1.04 for level or flat road
Speed V₁ = 60mil/hr = (60×5280)/(1×60×60) = 316800ft/3600s = 88ft/s
now, we substitute in our values to get the braking efficiency;
180ft = (1.04( (88ft/s)² - 0²)) / ( 2(32.2( (ηb/100)(0.80) + (0.018) ± sin(0°)))
180ft = 8053.76 / ( 64.4)(0.008ηb + 0.018)
180ft = 8053.76 / ( 0.5152ηb + 1.1592)
180( 0.5152ηb + 1.1592) = 8053.76
( 0.5152ηb + 1.1592) = 8053.76 /180
0.5152ηb + 1.1592 = 44.7431
0.5152ηb = 44.7431 - 1.1592
0.5152ηb = 43.5839
ηb = 43.5839 / 0.5152
ηb = 84.596 ≈ 84.6 %
Therefore, the unloaded braking efficiency is 84.6 %
Answer:
Explanation:
This is quite tricky! You need to do 2 different equations to solve all the parts of this problem. First is finding the acceleration in one dimension, which has an equation of
F - f = ma
where F is the applied Friction,
f is the frictional force acting against F,
m is the mass of the object, and
a is the acceleration of the object (NOT the velocity!)
This is Newton's Second Law expanded on a bit. The sum of the forces working on an object is equal to the object's mass times its acceleration. We have F, but we need f which is found in the equation
f = μ which is the coefficient of kinetic friction times the weight of the object. Weight is found in the equation
w = mg where m is mass and g is the pull of gravity. Let's start there and work backwards:
w = 37(9.8) to 2 sig figs so
w = 360N. Now fill that in to find f:
f = (.17)(360) to 2 sig figs so
f = 61. Now for the final answer in the original equation way back up at the top:
85 - 61 = 37a and do the subtraction on the left side first:
24 = 37a and then we divide to 2 sig figs to get
a = .65 m/s/s
Since we are moving in a straight line (as opposed to on an angle) the displacement is found in
d = rt which simply says that the distance an object moves is equal to its rate times the time. Therefore,
d = 2.2(3.4) to 2 sig figs so
d = 7.5 m