Answer:
the small mass will drop 11.65 cm below before come to rest
Explanation:
Given that:
the mass of the block (M) = 1.40 kg
the inclined angle θ = 25°
spring constant (k) = 30.0 N/m
suspended mass from the block (m) = 40.0 g = 0.040 kg
downward speed (v) = 1.20 m/s
The tension acting vertical on the inclined plane can be expressed as:
[tex]T = Mgsin 25^0 + kx - Ma \ \ \ ------equation (1)[/tex]
For the smaller mass (m)
[tex]T - mg = ma\\T = ma + mg \ \ \ ------ equation (2)\\[/tex]
If we equate equation (1) and (2) together; we have:
[tex]Mgsin 25^0 + kx - Ma = ma + mg[/tex]
making acceleration (a) the subject of the formula ; we have:
[tex]a = \frac{Mgsin25^0+kx-mg}{M+m} -------- equation (3)[/tex]
From the third equation of motion
[tex]v^2 =u^2 + 2as\\\\0 = (1.20)^22(-a)x\\\\a =\frac{(1.20)^2}{2x}\\\\a = \frac{0.72}{x}[/tex]
Replacing a with [tex]\frac{0.72}{x}[/tex] in equation (3); we have:
[tex]\frac{0.72}{x} = \frac{Mgsin25^0+kx-mg}{M+m}\\\\\\kx^2 +(Mgsin25^0 -mg)x = 0.72(M+m)[/tex]
From the equation above; let's substitute our given values;
Then. we have :
[tex]30x^2 +(1.40*9.8*0.4226 -0.040*9.8)x = 0.72(1.40+0.040)[/tex]
[tex]30x^2 +5.406x = 1.0368[/tex]
[tex]30x^2 +5.406x -1.0368[/tex] = 0
Using quadratic formula:
[tex]\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
where: a = 30 ; b = 5.406 ; c = -1.0368
Then;
[tex]\frac{-5.406 + \sqrt{5.3406^2-4(30)(-1.0368)}}{2*30} \ \ \ \ OR \ \ \ \frac{-5.406 - \sqrt{5.3406^2-4(30)(-1.0368)}}{2*30}[/tex]
= 0.1165 OR - 0.2967
The distance of how far it drops much be positive; so taking into account of the positive integer; x = 0.1165 m
x= 11.65 cm
Therefore; the small mass will drop 11.65 cm below before come to rest
