at the monkeys end:
monkey mass=W/g
tension=(w/g)(g+a)
but d=1/2 a t^2 or a=16*2/4=8 check that
so tension=W/g)(32+8)=40W/32=5W/4
but since tension = bananas
monkey=5W/4
Please reference the illustration at screenshots<dot>firefox<dot>com/fDDGSaztdbf0b7Uh/brilliant<dot>org
for the following problem:
A bunch of bananas of total weight W is hung at one end of a string passing over a perfectly smooth pulley. At the other end, a monkey starts climbing up the rope at a constant acceleration and covers 16 ft in 2 seconds.
If the bananas always remain at rest, find the weight of the monkey.
Assume that g=32 ft/sec²
Choices:
W/5
2W/5
4W/5
W
5W/4
Thank you.
2 answers
m = mass of monkey
M = mass of bananas
w = weight of monkey
W = weight of bananas
w = m g
W = M g
distance traveled = s
initial velocity = u
acceleration of monkey climbing = a
time = t
SUVAT equations of motion:
s = u t + a t² / 2
16 = 0 ∙ 2 + a ∙ 2² / 2
16 = 0 + a ∙ 4 / 2
16 = 2 a
2 a = 16
a = 16 / 2
a = 8 ft/s²
Since the bananas are stationary, by Newton's third law:
m ( g + a ) = M g
m ( 32 + 8 ) = M ∙ 32
40 m = 32 M
m = 32 M / 40 = 8 ∙ 4 / 8 ∙ 5 = 4 / 5 M
m = ( 4 / 5 ) M
w = W
m g = M g
m g = ( 4 / 5 ) M g
w = ( 4 / 5 ) W
M = mass of bananas
w = weight of monkey
W = weight of bananas
w = m g
W = M g
distance traveled = s
initial velocity = u
acceleration of monkey climbing = a
time = t
SUVAT equations of motion:
s = u t + a t² / 2
16 = 0 ∙ 2 + a ∙ 2² / 2
16 = 0 + a ∙ 4 / 2
16 = 2 a
2 a = 16
a = 16 / 2
a = 8 ft/s²
Since the bananas are stationary, by Newton's third law:
m ( g + a ) = M g
m ( 32 + 8 ) = M ∙ 32
40 m = 32 M
m = 32 M / 40 = 8 ∙ 4 / 8 ∙ 5 = 4 / 5 M
m = ( 4 / 5 ) M
w = W
m g = M g
m g = ( 4 / 5 ) M g
w = ( 4 / 5 ) W