Question

(a) A 1/2 kg book is shoved across the table with an initial velocity of 2 m/s. If it slides for a 1 meter before coming to a rest, (a) What was its acceleration? (Technically, this acceleration is negative, since the book is slowing down.) (b) What was the frictional force?

(b) A "synchronous" satellite is one that remains above the same point on the equator at all times. (a) What is the period of a synchronous satellite? (b) How high above the surface must it be?

Answer 1

1A.

Using 3rd kinematic equation:

V^2 = U^2 + 2*a*d

d = stopping distance = 1 m

a = acceleration of book = ?

U = Initial speed of book = 2 m/s

V = final speed = 0 m/s

So,

a = (V^2 - U^2)/(2*d) = (0^2 - 2^2)/(2*1) = -2 m/s^2

acceleration of book = -2 m/s^2 (So we can see that acceleration is negative)

Magnitude of acceleration = 2 m/s^2

1B.

Since there is only one force on the book and that is friction force (Since book is slowing down only because of friction force), So

F_net = F_fric

F_fric = m*a

m = mass of book = 1/2 kg

So,

F_fric = (1/2)*(-2) = -1 N

Friction force = -1 N

2A.

Since "synchronous" satellite is one that remains above the same point on the equator at all times, So It's period will be equal to one days, same as earth

So Period of synchronous satellite = 1 day = 86400 sec

2B.

Using force balance on sattelite

Centripetal froce = Gravitational force

Fc = Fg

mv^2/r = GMm/r^2

r = Re + h

v = w*r

w = 2*pi/T

m*w^2*r = GMm/r^2

4*pi^2/T^2 = GM/r^3

(Re + h)^3 = GMT^2/4*pi^2

h = [GMT^2/(4*pi^2)]^(1/3) - Re

Re = 6.371*10^6 = Radius of earth

M = 5.98*10^24 kg = Mass of earth

T = 24 hr = 24*3600 sec = time period

h = [6.67*10^-11*5.98*10^24*(86400)^2/(4*pi^2)]^(1/3) - 6.371*10^6

h = 35879.474 km = 3.59*10^7 m

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