Answer:
Tangential speed=5.4 m/s
Radial acceleration=
Explanation:
We are given that
Angular speed=2.59 rev/s
We know that
1 revolution=
2.59 rev=
By using 
Angular velocity=
Distance from axis=r=0.329 m
Tangential speed=
Radial acceleration=
Radial acceleration=
A 300-kg bear grasping a vertical tree slides down at constant velocity. The friction force between the
tree and the bear is
Answer:
(a) work required to lift the object is 1029 J
(b) the gravitational potential energy gained by this object is 1029 J
Explanation:
Given;
mass of the object, m = 35 kg
height through which the object was lifted, h = 3 m
(a) work required to lift the object
W = F x d
W = (mg) x h
W = 35 x 9.8 x 3
W = 1029 J
(b) the gravitational potential energy gained by this object is calculated as;
ΔP.E = Pf - Pi
where;
Pi is the initial gravitational potential energy, at initial height (hi = 0)
ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)
ΔP.E = 1029 J
