Answer:
(a) 1.054 m/s²
(b) 1.404 m/s²
Explanation:
0.5·m·g·cos(θ) - μs·m·g·(1 - sin(θ)) - μk·m·g·(1 - sin(θ)) = m·a
Which gives;
0.5·g·cos(θ) - μ·g·(1 - sin(θ) = a
Where:
m = Mass of the of the block
μ = Coefficient of friction
g = Acceleration due to gravity = 9.81 m/s²
a = Acceleration of the block
θ = Angle of elevation of the block = 20°
Therefore;
0.5×9.81·cos(20°) - μs×9.81×(1 - sin(20°) - μk×9.81×(1 - sin(20°) = a
(a) When the static friction μs = 0.610 and the dynamic friction μk = 0.500, we have;
0.5×9.81·cos(20°) - 0.610×9.81×(1 - sin(20°) - 0.500×9.81×(1 - sin(20°) = 1.054 m/s²
(b) When the static friction μs = 0.400 and the dynamic friction μk = 0.300, we have;
0.5×9.81·cos(20°) - 0.400×9.81×(1 - sin(20°) - 0.300×9.81×(1 - sin(20°) = 1.404 m/s².
Interactúan dos objetos cargados dependiendo de la carga domínate en el objeto, ya que cargas iguales se repelen y cargas opuestas se atraen. ... Al frotar un cuerpo cargado con otro objeto de carga neutra, este se cargara pero adquiriendo la carga opuesta al objeto cargado inicialmente.
Answer
given,
Initial speed of the car = 0 m/s
acceleration of car = 2.4 m/s²
Speed of the truck = 15.5 m/s
to car to overtake the truck distance travel by both should be same
distance traveled by the truck
d = 15.5 t
distance traveled by the car
d = u t + 1/2 gt ²
d = 0.5 x 9.8 t²
now, equating both the equations
15.5 t = 4.9 t²
t = 3.163 s
Time taken by car to overtake truck is 3.163 s
distance travel by the car = 15.5 x 3.163
d = 49 m
b) Speed of the automobile
using equation of motion
v = u + at
v = 0 + 2.4 x 3.163
v = 7.59 m/s
Speed of the automobile while crossing truck is equal to v = 7.59 m/s
