Two blocks with different mass are attached to either end of a light rope that p
ID: 1537204 • Letter: T
Question
Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended a distance 1.40 m , its speed is 3.00 m/s .
Part A
If the total mass of the two blocks is 16.0 kg , what is the mass of the more massive block?
Take free fall acceleration to be 9.80 m/s2 .
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Part B
What is the mass of the lighter block?
Take free fall acceleration to be 9.80 m/s2 .
SubmitMy AnswersGive Up
Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended a distance 1.40 m , its speed is 3.00 m/s .
Part A
If the total mass of the two blocks is 16.0 kg , what is the mass of the more massive block?
Take free fall acceleration to be 9.80 m/s2 .
m1 = kgSubmitMy AnswersGive Up
Part B
What is the mass of the lighter block?
Take free fall acceleration to be 9.80 m/s2 .
m2 = kgSubmitMy AnswersGive Up
Explanation / Answer
Given
two masses are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended a distance 1.40 m , its speed is 3.00 m/s .
and kg
the sum of the masses m1+m2 = 16
let mass m1 is descending and m2 ascending
we knwo the acceleration of the system is a = g (m1-m2)/(m1+m2) -----(1)
using equations of motions V^2 - u^2 = 2as
3^2 - 0 = 2*a*1.4
a = 3.21 m/s2
from (1) 3.21 = 9.8(m1-m2)/16 ===> m1-m2 = 5.24 kg
==> m1 = 5.24 +m2
m1+m2 = 16 kg
now 16 - (5.24+m2) = m2
2*m2 = 16-5.24
m2 = (16-5.24)/2 kg
m2 = 5.38 kg
so
a) m1 = 5.24+5.38 = 10.62 kg
b) m2 = 5.38 kg
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