A 12.0 kg monkey starts to climb a long rope to reach a banana located at a height of 6.95 m. The rope will snap if the tension exceeds 132.0 N. Calculate the least amount of time the monkey could take to reach the banana without breaking the rope.
12 kg is 117.6 N, so the difference, 132–117.6 = 14.4 N. That is the max force he can apply to the rope.
F = ma
14.4 = 12a
a = 1.2 m/s²
assuming the 6.95 m number includes the height of the monkey, so he has to move that far up the rope.
d = ½at²
6.95 = ½(1.2)t²
t = 3.4 seconds
.
January 26th, 2010 at 12:27 am
12 kg is 117.6 N, so the difference, 132–117.6 = 14.4 N. That is the max force he can apply to the rope.
F = ma
14.4 = 12a
a = 1.2 m/s²
assuming the 6.95 m number includes the height of the monkey, so he has to move that far up the rope.
d = ½at²
6.95 = ½(1.2)t²
t = 3.4 seconds
.
References :
January 26th, 2010 at 1:04 am
The monkey begins by hanging (at rest) at the bottom of the rope. In this situation, he will exert a force (tension) on the rope of
T = mg
where m is his mass and g is the acceleration due to gravity. To climb, he must accelerate and if he’s accelerating upwards at a m/s, the tension becomes
T = m(g+a)
Solving this for a gives
a = (T/m) – g
Now, starting from rest, the distance, d, traveled with constant acceleration, a, is
d = 1/2 a t^2
so
t = sqrt(2d/a)
Plugging in a from above we have
t = sqrt( 2d / [(T/m) - g] )
With d = 6.95 m, T = 132 N, m = 12 kg, and g = 9.8 m/s^2, we get
t = 11.58 s
References :