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Frictional force = µN
µ = 1 ÷ 2√3
N = mg*cosø
g = 10 m/s/s
cosø = √3/2
N = 5√3m
frictional force = 2.5m N
Force after brakes are applied = mg
= 10m N
Net force = 7.5m N
force = ma
7.5m = ma
-> a = 7.5 m/s/s
initial velocity = 21.6 km/hr
= 6 m/s
distance of the slope = 12.8m
3rd equation of motion:
v^2 = u^2 + 2as
where u = initial velocity and v = final velocity
v^2 = 36 + 192
= 228
-> v = √228
36 km/hr = 10 m/s
√228 > 10 because √228 square = 228 and 10 square = 100.
µ = 1 ÷ 2√3
N = mg*cosø
g = 10 m/s/s
cosø = √3/2
N = 5√3m
frictional force = 2.5m N
Force after brakes are applied = mg
= 10m N
Net force = 7.5m N
force = ma
7.5m = ma
-> a = 7.5 m/s/s
initial velocity = 21.6 km/hr
= 6 m/s
distance of the slope = 12.8m
3rd equation of motion:
v^2 = u^2 + 2as
where u = initial velocity and v = final velocity
v^2 = 36 + 192
= 228
-> v = √228
36 km/hr = 10 m/s
√228 > 10 because √228 square = 228 and 10 square = 100.
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