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.