What is the required force to bring a 1550 kg vehicle moving at 32 m/s to a stop in a distance of 45 m?
1 answer:
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at the lowest point in the trajector, the kinetic energy of the bob is 300 J.
Explanation:
The total mechanical energy of the bob at any point of its motion is given by
Where
is the kinetic energy, where
m is the mass of the bob
v is its speed
is the gravitational potential energy, where
g is the acceleration of gravity
h is the height of the bob, measured with respect to the lowest point of the trajector
In absence of friction, the total mechanical energy E remains constant. So we have:
- When the bob swings upward, the PE increases (because h increases) and the KE decreases (so the speed decreases). At the highest point in the trajector, the speed of the bob is zero (v=0), so its KE is also zero and all the mechanical energy is potential energy: U = 300 J
- When the bob swings downward, the PE decreases (because h decreases) and the KE increases (so the speed increases). At the lowest point in the trajectory, the height has become zero (h=0), so the PE is zero and all the mechanical energy is kinetic energy: KE = 300 J
Therefore, at the lowest point in the trajector, the kinetic energy of the bob is 300 J.
Learn more about kinetic energy:
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Answer:
The minimum speed must the car must be 13.13 m/s.
Explanation:
The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.
We know that, mg be the weight of car and rider, which is equal to the centripetal force.
So, the minimum speed must the car must be 13.13 m/s.