How Far Will the Mass Travel Beyond the Incline After Being Released?

[312213]

Homework Statement

A spring of length 0.80 m rests along a frictionless 30° incline (a). A 2.6 kg mass, at rest against the end of the spring, compresses the spring by 0.10 m. (b)
http://img64.imageshack.us/img64/5889/physics.jpg

The mass is pushed down, compressing the spring an additional 0.60 m, and then released. If the incline is 2.0 m long, determine how far beyond the rightmost edge of the incline the mass lands.

The spring constant is 127.4

Homework Equations

Us=1/2kx²
v²=v0²+2ax
v=v0+at
x=x0+v0t+1/2at²

The Attempt at a Solution

0.8m, in a 30° angle, on the spring will count as the 0 point.
The potential energy of the spring on the block is 1/2kx²=U_s so
(1/2)(127.4)(0.1+0.6)²=31.213J

K_i+U_i=K_f+U_f
0+31.213=K_f+0

K=1/2mv²
31.213=1/2(2.6)v²
4.9=v

http://img691.imageshack.us/img691/5889/physics.jpg
Diagram of forces show gravity acts against it so mgsin30
(2.6)(-9.8)sin30=-12.74N
F=ma
-12.74=2.6a
-4.9=a

v²=v₀²+2a(x-x₀)
v²=4.9²+2(-4.9)(1.9)
2.32=v

So the block leaves the ramp at ~2.32m/s, 30° above the horizontal.

I use the time it takes for the block to reach its peak height and time to hit the ground and multiply that with the velocity it moves rightwards.

v=v₀+at
0=2.32sin30+-9.8t
0.12=t1

In this time the block moves up so its final height is
x=x₀+v₀t+1/2at²
x=0+(2.32sin30)0.12+(1/2)(-9.8)(0.12)²
x=0.06875+1 (original height of right edge of ramp)
x=1.06875

Next is the time it takes to land,
x=x₀+v₀t+1/2at²
1.06875=0+0+(1/2)(9.8)t²
0.467=t2

t1+t2=0.585

Finally the distance it goes away from the edge is velocity times time so

x=vt
x=(2.32cos30)(0.585)
x=1.1771557879568892088679537911192

Is this right or are there something wrong with the way I approached the problem?

 

[Doc Al]

The Attempt at a Solution

0.8m, in a 30° angle, on the spring will count as the 0 point.
The potential energy of the spring on the block is 1/2kx²=U_s so
(1/2)(127.4)(0.1+0.6)²=31.213J

That's the elastic potential energy at the lowest point.

K_i+U_i=K_f+U_f
0+31.213=K_f+0

Don't forget gravitational potential energy.

Hint: Rather than treating the spring and gravity separately, use conservation of energy to solve for the speed of the mass as it leaves the incline.

 

[kerimek]

Your approach and calculations seem correct. However, when determining the distance the block travels beyond the rightmost edge of the incline, it is important to consider the horizontal displacement caused by the initial compression of the spring (0.10m). This can be calculated using the equation x=vt, where v is the horizontal velocity of the block (2.32cos30) and t is the time it takes for the block to reach its peak height (0.12s). This would result in an additional horizontal displacement of 0.278m. Therefore, the total distance the block travels beyond the rightmost edge of the incline is 1.455m.