What kind of variation is there in the mechanical energy as the cart rolls down the ramp? Does this agree with your prediction?
1 answer:
Answer:
Em₀ = U = m g h , Em_{f} = K = ½ m v²
Explanation:
When a car is on a ramp it has a certain amount of mechanical energy. At the highest point of the ramp the mechanical energy is fully potential given by
Em₀ = U = m g h
As part of this energy descends down the ramp, part of this energy is transformed into kinetic energy and has one part of each, even though the sum remains the initial energy
Em = K + U = ½ m v² + mg y
y <h
when it reaches the bottom of the ramp it has no height therefore there is no potential energy, all of it has been transformed into kinetic energy
Em_{f} = K = ½ m v²
This energy transformation is in the case that the friction force is zero.
If there is a friction force, it performs work against the low car, it is reflected in an increase in the internal energy (temperature) of the car. In this case the energy in the lower part is less than the initial one by a factor
= - fr L
therefore the numeraire values of the velocity are lower, due to the energy lost by friction.
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Answer:
They will move the fridge if they all push in the same direction, but it will not move with constant velocity
Explanation:
The maximum static friction force is
(negative sign since its direction is opposite to the push applied by the people)
Sam can apply a force of 130 N, while Amir and Andre can apply a push of 65 N each, so the total force that they can apply, if they push in the same direction, will be:
This force is larger than the frictional force, so the fridge will start moving.
However, the net force on the fridge will be:
And according to Newton's second law,
where m is the mass of the fridge and a its acceleration, since the net force is not zero, then the fridge will have a non-zero acceleration, so it will not move with constant velocity.