A spacecraft flies away from the earth with a speed of 4.80×106m/s relative to the earth and then returns at the same speed. The
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(a) 73.5 N
The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law
the resultant of the forces must be zero: (1)
The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:
, the push applied by the worker
, the force of friction, with
being the coefficient of friction,
being the mass of the crate, and
. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes
So, this is the force that the worker must apply.
(b) 330.8 J
The work done by the pushing force of the worker on the crate is given by:
where
F = 73.5 N is the force
d = 4.5 m is the displacement
is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)
Substituting, we have
(c) -330.8 J
To calculate the work done by friction, we apply the same formula:
where
is the magnitude of the force of friction
d = 4.5 m is the displacement
is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)
Substituting, we find
So, the work done by friction is negative.
(d) 0 J
As before, the work done by any force on the crate is
We notice that both gravity and normal force are perpendicular to the displacement: therefore, , and so
which means that the work done by both forces is zero.
(e) 0 J
The total work done on the crate is the sum of the work done by the four forces acting on it, so:
And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.