Answer:
20.25 m
Explanation:
- <u>Centripetal acceleration </u>is given by; the square of the velocity, divided by the radius of the circular path.
That is;
<em><u>ac = v²/r</u></em>
<em> </em><em><u> Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the radius, m</u></em>
Therefore;
r = v²/ac
= 27²/36
= 20.25 m
Hence the radius is 20.25 meters
<span>pile
battery
<span>power sector</span></span>
When the box IS on the shelf 2m above the ground,
its potential energy is
(weight) x (height) = (3 N) x (2 m) = 6 joules .
THAT's the work you have to do, to lift the box up to there.
Since the ladder is standing, we know that the coefficient
of friction is at least something. This [gotta be at least this] friction
coefficient can be calculated. As the man begins to climb the ladder, the
friction can even be less than the free-standing friction coefficient. However,
as the man climbs the ladder, more and more friction is required. Since he
eventually slips, we know that friction is less than what's required at the top
of the ladder.
The only "answer" to this problem is putting lower
and upper bounds on the coefficient. For the lower one, find how much friction
the ladder needs to stand by itself. For the most that friction could be, find
what friction is when the man reaches the top of the ladder.
Ff = uN1
Fx = 0 = Ff + N2
Fy = 0 = N1 – 400 – 864
N1 = 1264 N
Torque balance
T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)
N2 = 439 N
Ff = 439= u N1
U = 440 / 1264 = 0.3481