Answer:
C) W = - 190 J
Explanation:
Notation
Wf = work done by the friction force (unknown)
Ff = force of the friction
d = distance travelled by the box = (2 pi 1.82 m) = 11.435 m
-GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.
Given
A particle of mass m moving under the influence of a fixed mass's M, gravitational potential energy of formula -GMm/r, where r is the separation between the masses and G is the gravitational constant of the universe.
As the Gravity Potential energy of particle = -GMm/r
Total energy of particle = Kinetic energy + Potential Energy
As we know that
Kinetic energy = 1/2mv²
Also, v is equals to square root of GM/r
v = √GM/r
Put the value of v in the formula of kinetic energy
We get,
Kinetic Energy = GMm/2r
Total Energy = GMm/2r + (-GMm/r)
= GMm/2r - GMm/r
= -GMm/2r
Hence, -GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.
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<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
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