A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground.
1 answer:
<h2>Thus the force of friction is 235 N</h2>
Explanation:
When the bear was at the height of 14 m . Its potential energy = m g h
here m is the mass of bear , g is acceleration due to gravity and h is the height .
Thus P.E = 27 x 10 x 14 = 3780 J
The K.E of the bear just before hitting = m v²
= x 27 x ( 6.1 )² = 490 J
The force of friction f = P.E - K.E = 3290 J
Because the work done = Force x Distance
Thus frictional force = = 235 N
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Answer:
18.7842493212 W
Explanation:
T = Tension = 1871 N
= Linear density = 3.9 g/m
y = Amplitude = 3.1 mm
= Angular frequency = 1203 rad/s
Average rate of energy transfer is given by
The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W