Respuesta :
Answer
given,
speed of the car = 50 Km/h
= 50 x 0.278 = 13.9 m/s
a) normal force acting on the car
[tex]F_N = m g[/tex]
force along x-direction
[tex]F_x = m a_x[/tex]
friction act along x-direction in opposite direction
[tex]-f_{friction} = m a_x[/tex]
[tex]-\mu F_N = m a_x[/tex]
[tex]-\mu m g= m a_x[/tex]
[tex]a_x= -\mu g[/tex]
a) using equation of motion
v² = u² + 2 a s
final speed, v = 0 m/s
0² = 13.9² - 2 x μg x s
2 x 0.1 x 9.8 x s = 13.9²
s = 98.57 m
minimum distance in which the car will stop is equal to 98.57 m
b) When coefficient of friction is 0.6
v² = u² + 2 a s
final speed, v = 0 m/s
0² = 13.9² - 2 x μg x s
2 x 0.6 x 9.8 x s = 13.9²
s = 16.43 m
stopping distance when coefficient of friction is 0.600 is equal to 16.43 m
