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

the small mass will drop 11.65 cm below before come to rest

Step-by-step 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 = (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 =((1.20)^2)/(2x)\\\\a = (0.72)/(x)`

Replacing a with
`(0.72)/(x)` in equation (3); we have:

`(0.72)/(x) = (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:

`(-b \pm √(b^2-4ac))/(2a)`

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

Then;

`(-5.406 + √(5.3406^2-4(30)(-1.0368)))/(2*30) \ \ \ \ OR \ \ \ (-5.406 - √(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

Step-by-step explanation:

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

Kindly check ATTACHED PICTURE for complete solution.