# Finding block flight distance (work/energy/projectile motion)

1. Oct 21, 2009

### Gotejjeken

1. The problem statement, all variables and given/known data

The spring in the figure has a spring constant of 1100 N/m. It is compressed 14.0 cm, then launches a 200 g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.210.

What distance d does the block sail through the air?

2. Relevant equations

Work of spring:

W(spring) = 1/2*k*s^2

Work of Friction + Gravity as block slides up incline:

W(friction) = -u*k*n*s = -u*k*Wx*s = -u*k*m*g*cos(theta) * s
W(gravity) = Wy * s = m * g * s * sin(theta)

Work-Energy Theorem

W(net) = (delta)K = Kf - Ki

Projectile Motion Kinematics:

Vfy = Viy - g * t
Xf = Xi + Vix * t

3. The attempt at a solution

Using the work equations above, I found the net work at the top of the ramp to be 10.47J (10.7J from spring initially - 0.041J of friction - 0.19J of gravity = 10.47J), and then I used the Work-Energy equation to find the final velocity at the top of the ramp to be 10.23 m/s.

Using this velocity and the fact that the launch angle is 45 degrees, I found the x and y components of the launch velocity and plugged them into the equations above to get the time the block spent in the air, and then finally the final distance traveled by the block, which I found to be 5.41m.

Unfortunately, this answer is incorrect. After looking around the web for a while for another reasonable solution I found that most answers given were about double mine, as in 10.48m instead of 5.41m. After trying this answer, I found it was also wrong.

Any help would be appreciated.

2. Oct 21, 2009

### mgb_phys

Looks like the correct method you have probably messed up the calculations somewhere.

You have PE as m g s sin(th), but you are given the vertical height.
Did you get the normal force right for the friction?

3. Oct 21, 2009

### Gotejjeken

I drew a Free Body Diagram along the ramp, so friction is pointing down the negative x-axis, the normal is perpendicular to the motion along the positive y-axis, and the weight is pointing downward at an angle of 45 in the fourth quadrant (below positive x-axis and to the right of the negative y-axis). This diagram leaves me with a component of weight in the x and y direction and the Fnet equations:

(Fnet)x: Wx - f = M * a
(Fnet)y: n - Wy = 0 => n = Wy => n = Wcos(theta) = > m * g * cos(45)

Then I found the work using those Fnet equations. Maybe I drew my FBD wrong?

4. Oct 21, 2009

### mgb_phys

You only need a FBD for the friction, remember frictional force acts along the surface (ie the hypotonuse of the triangle in this case) and F = coef * force.NORMAL to the surface

For the PE you are probably making it too complicated.
The vertical height is 2m so the PE is just = 0.2 * 9.8 *2 (the slope doesn't matter)

5. Oct 21, 2009

### Gotejjeken

So the work done by friction would be 4.12J in that case, leaving me with of a Wnet of 10.70J - 4.12J = 6.58J, correct? From there I would just get the velocity as before and plug it into those kinematic equations to get distance?

If I do that, I get a new distance of 3.32m, although I'm not entirely sure if it should be smaller than before. I might still be messing up the calculations somewhere...

6. Oct 21, 2009

### mgb_phys

Length of slope = 2m / cos(45) = 2.828m (hint cos(45) = sqrt(2)/2 )
Normal force = m g cos(45) = 0.1414 * 9.8N = 1.4N

So energy lost to friction = 0.21 * 1.4N * 2.83m = 0.82J

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