(a) magnitude of acceleration:
we will imagine that the two components of the force are the sides of a right-angled triangle.
We can thus use the Pythagorean theorem to find the magnitude of the force which is the hypotenuses of this triangle
Magnitude of force = sqrt(390^2 + 180^2) = 429.53 Newton,
Using Newton's second law (F=ma), we can find the acceleration as follows:
429.53 = 270 * a
a = 1.59 m/sec^2
(b) direction of acceleration:
we will again consider the right angled triangle formed from the components of the force.
Using the trigonometric function (tan = opposite / adjacent), we can find the direction of the force as follows:
tan theta = 390 / 180
tan theta = 2.1667
theta = 65.22 degrees
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Explanation:
Potential energy = mgh
So, energy gained
= mgh
= 70kg × (9.8m/s²) × 1000m
= 686000 kgm²/s²
= 686000 J
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,
⇒