Explanation:
It took for the sound to reach the 1st wall and at the same time time, the same sound took
to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let
be the distance of the person to the 1st wall and
be the distance to the 2nd wall. So the distance between the walls X is
Answer:
the mass drop by 6.5cm before coming to rest.
Explanation:
Given that:
the mass of the block M= 1.40 kg
angle of inclination θ = 30°
spring constant K = 40.0 N/m
mass of the suspended block m = 60.0 g = 0.06 kg
initial downward speed = 1.40 m/s
The objective is to determine how far does it drop before coming to rest?
Let assume it drops at y from a certain point in the vertical direction;
Then :
Workdone by gravity on the mass of the block is:
= - 6.867y
Workdone by gravity on the mass of the suspended block is:
= 0.5886
The workdone by the spring = -1/2ky²
= -0.5 × 40y²
= -20 y²
The net workdone is = -20 y² - 6.867y + 0.5886y
According to the work energy theorem
Net work done = Δ K.E
-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²
-20 y² - 6.867y + 0.5886 = 0.5 × 0.1176
-20 y² - 6.867y + 0.5886 = 0.0588
-20 y² - 6.867y + 0.5886 - 0.0588
-20 y² - 6.867y + 0.5298 = 0
multiplying through by (-)
20 y² + 6.867y - 0.5298 = 0
Using the quadratic formula:
where;
a = 20 ; b = 6.867 c= - 0.5298
= 0.0649 OR −0.408
We go by the positive integer
y = 0.0649 m
y = 6.5 cm
Therefore; the mass drop by 6.5cm before coming to rest.