An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ abov
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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