An object with an initial speed of 4.0 m/s accelerates uniformly at 2.0 m/s^2
1 answer:
You might be interested in
Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as
= velocity of car A before collision = 0 i -
j
= velocity of car B before collision =
i + 0 j
= velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j
= mass of car A = 1750 kg
= mass of car B = 1450 kg
Using conservation of momentum
+
= (
+
) (
)
(1750) (0 i - j) + (1450) (
i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)
(1450) i - (1750)
j = 97600 i - 60160 j
Comparing the coefficient of "i" and "j" both side
(1450) = 97600 and - (1750)
= - 60160
= 67.3 km/h and
= 34.4 km/
Answer:
μs = 0.75
μk = 0.58
Explanation:
From a force diagram:
(1)
(2)
When it starts slipping, friction force is the maximum and acceleration is 0. Replacing these conditions on (1):
Solving for μs:
μs = tan 36.7° = 0.75
When it moves at constant speed, friction force is kinetic friction and acceleration is 0. With these conditions the coefficient is:
μk = tan 30.1° = 0.58