2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is
applied to the box. The force is directed to the left and has magnitude F1t2 = 16.00 N>s22t2. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?
2 answers:
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Answer:
Explanation:
The total energy of the satellite when it is still in orbit is given by the formula
where
G is the gravitational constant
m = 525 kg is the mass of the satellite
is the Earth's mass
r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:
So the initial total energy is
When the satellite hits the ground, it is now on Earth's surface, so
so its gravitational potential energy is
And since it hits the ground with speed
it also has kinetic energy:
So the total energy when the satellite hits the ground is
So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:
The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.