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
You might be interested in
Answer:
Resistance of the iron rod, R = 0.000077 ohms
Explanation:
It is given that,
Area of iron rod,
Length of the rod, L = 35 cm = 0.35 m
Resistivity of Iron,
We need to find the resistance of the iron rod. It is given by :
So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.