The decrease of PE for a freely falling object equals its gain in KE, in accord with the conservation of energy. By simple algeb
1 answer:
Answer:
Explanation:
The potential energy (PE) we are looking here is gravitational potential energy (GPE).
GPE= mgh,
where m is the mass of an object,
g is the gravitational field strength
h is the height of the object
KE= ½mv²,
where m is the mass and v is the velocity
loss in GPE= gain in KE
mgh= ½mv²
gh= ½v² (<em>divide by m throughout</em>)
Assuming that the object is on earth, then g= 10N/Kg
½v²= 10h (<em>substitute g=10</em>)
v²= 20h (<em>×2 on both sides</em>)
v= √20h (<em>square root both sides</em>)
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Answer:
Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W
Explanation:
The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.
The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.
then, the work done by the engine is:
The formulas for the Kinetic and potential energy are:
where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.
Using the formulas:
Replacing the values:
The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.
The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:
the power delivered by the car's engine was: