find the initial separation L between mass and spring ?

Question

a 2200 gram mass starts from rest and slides a distance L down a frictionless 40 degree incline, where it contacts an unstressed 60 cm long spring of negligible mass. the mass slides an additional 20 cm as it is brought momentarily to rest by compressing the spring of force constant 13 n/cm. the acceleration of gravity is 9.8 m/s^2. note: the spring lies along the surface of the ramp. the ramp is frictionless. the external force is rapidly removed so that the compressed spring can push up the mass. find the initial separation L between mass and spring ?

Answer 1

Ok here we have a plane incline. The angle of the slope is 40º. There is no friction, so we can use the conservation of energy.

the mass is : 2.2 kg

the constant "K" of the spring : 13*100 N/m

And the mass starts from an altitude : H

We just need to find the altitude, because if you make a graphic, you will find a right triangle, the hypotenuse will :

L + (20/100) m

gravity = 9.8 m/s^2

So here we go, conservation of energy :

2.2*9.8*H = 1/2*1300*(20/100)^2

H = 1.2 meters

So we have the altitude : 2.2 m

Now I told you, that you will find a right triangle, using the 40º angle :

(L + (20 / 100))sin(40) = H = 2.2

L + 0.2 = 3.42 >>>> L = 3.22 m

Here is the explanation, the spring compresses 20 cm, I used meters, (20 / 100), the lenght of the hypotenuse is :

L + (20 / 100), and with the value of H, I used trigonometry.

Hope that might help you

Answer 2

From conservation of energy (in a conservative force field - gravitational):

mg(L+.2)sin40 = 1300 N/m * (.2m)^2

The left side is the gravitational potential energy of the mass m before it slides down the frictionless incline.

The right side is the elastic energy stored in the spring after it's compressed by .2 m (= 20 cm).

One linear equation and one unknown - easy to solve for L.

L = [ 1300 N/m * .04 m^2 ] /[2.2 kg * 9.8m/s^2 * sin(40)] - 0.2m
= 52 / 13.86 m - 0.2 m = 3.6 m

Note: The mass slides down the incline 3.6 m, but it only "falls" a distance of (L + .2)*sin(40). The horizontal component of motion down the incline doesn't add to it's potential energy.

Answer 3

i do no longer evaluate like doing this being a accurate college pupil on summer season time vacation: inspite of the actuality that, i'd be able to allow you to know its some style of share with time being a consistent... stress=mass x acceleration a.ok.a. 3.02m/2.73kg=2.55m/Xkg some issue like that..... if now no longer then oh sturdy.