Question

A 1.40-kg block is on a frictionless, 25 ∘ inclined plane. The block is attached to a spring (k = 30.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 40.0-g suspended mass. The suspended mass is given an initial downward speed of 1.20 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)

Answer 1

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:

$T = Mgsin 25^0 + kx - Ma \ \ \ ------equation (1)$

For the smaller mass (m)

$T - mg = ma\\T = ma + mg \ \ \ ------ equation (2)$

If we equate equation (1) and (2) together; we have:

$Mgsin 25^0 + kx - Ma = ma + mg$

making acceleration (a) the subject of the formula ; we have:

$a = \frac{Mgsin25^0+kx-mg}{M+m} -------- equation (3)$

From the third equation of motion

$v^2 =u^2 + 2as\\\\0 = (1.20)^22(-a)x\\\\a =\frac{(1.20)^2}{2x}\\\\a = \frac{0.72}{x}$

Replacing a with $\frac{0.72}{x}$ in equation (3); we have:

$\frac{0.72}{x} = \frac{Mgsin25^0+kx-mg}{M+m}\\\\\\kx^2 +(Mgsin25^0 -mg)x = 0.72(M+m)$

From the equation above; let's substitute  our given values;

Then.  we have :

$30x^2 +(1.40*9.8*0.4226 -0.040*9.8)x = 0.72(1.40+0.040)$

$30x^2 +5.406x = 1.0368$

$30x^2 +5.406x -1.0368$ = 0

Using quadratic formula:

$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

where: a = 30 ; b = 5.406 ; c = -1.0368

Then;

$\frac{-5.406 + \sqrt{5.3406^2-4(30)(-1.0368)}}{2*30} \ \ \ \ OR \ \ \ \frac{-5.406 - \sqrt{5.3406^2-4(30)(-1.0368)}}{2*30}$

= 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

Answer 2

Answer: The small mass will drop below 11.6cm before coming to rest

Explanation:

Equating the Tension of both the two masses, and inputting the given parameters adequately...

Kindly check ATTACHED PICTURE for complete solution.