Question

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.

Answer 1

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

Answer 2

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