A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca
r from sliding down the incline.
1 answer:
Answer:
Force of friction, f = 751.97 N
Explanation:
it is given that,
Mass of the car, m = 1100 kg
It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.
From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.
f = 751.97 N
So, the force of friction on the car is 751.97 N. Hence, this is the required solution.
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Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,
Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀ =A₀/8
⇒
applying logarithm on both sides
⇒
⇒
substituting the values
b)
, where
is time period of damped SHM
⇒
let be number of oscillations made
then,
⇒