# Work Energy Theorem and Kinematics

1. Mar 8, 2014

### layout4it

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

A small steel ball of mass .0283kg is placed on the end of a plunger of length .0051m attached to a spring 1.88m above the ground. The spring is pre-compressed .0011m and has a spring constant of 177 N/m. The plunger is then angled on a ramp 45° above the horizontal, and is pressed in to compress the spring an additional .0088m. The plunger is then released extending to the end of the ramp and sending the ball into the air. Assuming no friction and no air resistance how far will the ball fly before hitting the ground?

2. Relevant equations

Kinematics
$Δx = v_0 t + 1/2 a t^2$
$v^2 = v_0^2 + 2aΔx$
$v = v_0 + at$

Work/Energy
$W = ΔK$
$KE= 1/2 m v^2$
$PEspring = 1/2 k Δx^2$
$PEgravity = mgh$

3. The attempt at a solution

I split the problem into 3 parts: Launch, End of Launch → Max Height, Max Height → Ground

Launch

$W = ΔK$
(no friction or air resistance)
$W=0$

$∴K_0 = K_f$

$KE_0 + PEgravity_0 + PEspring_0 = KE_f + PEgravity_f + PEspring_f$

$1/2 m v_0^2 + mgh_0 + 1/2 k Δx_0^2 = 1/2 m v_f^2 + mgh_f + 1/2 k Δx_f^2$

Calling $h_0$ the ground

$0 + mgh_0 + 1/2 k Δx_0^2 = 1/2 m v_f^2 + mgh_f + 1/2 k Δx_f^2$

Multiply both sides by 2 to get rid of the fractions

$2mgh_0 + k Δx_0^2 = m v_f^2 + 2mgh_f + kΔx_f^2$

Bring Like Terms Together

$k Δx_0^2 - kΔx_f^2 = mv_f^2 + 2mgh_f - 2mgh_0$

Factor Out Mass

$k Δx_0^2 - kΔx_f^2 = m(v_f^2 +2gh_f - 2gh_0)$

Divide Both Sides By Mass

$\frac{(k Δx_0^2 - kΔx_f^2)}{m} = v_f^2 +2gh_f - 2gh_0$

Isolate $V_f$

$\frac{(k Δx_0^2 - kΔx_f^2)}{m} +2gh_0 - 2gh_f = v_f^2$

Solve for $V_f$
$v_{flight} = \sqrt{\frac{(k Δx_0^2 - kΔx_f^2)}{m} +2gh_0 - 2gh_f }$
(I name it $v_{flight}$ for simplicity)

End of Launch → Max Height

$v_{0y} = v_{flight} \sin{45°}$

$v_{fy} = v_{0y} + a_y t$

$t_1 = \frac {v_{fy} - v_{0y}} {a_y}$

$t_1 = \frac {0 - v_{0y}} {a_y}$

$t_1 = \frac {-v_{0y}} {a_y}$

Max Height → Ground

$v_{fy}^2 = v_{0y}^2 + 2a_yΔy$

$v_{fy}^2 = v_{0y}^2 + 2a_yΔy$

$v_{fy}^2 = 0 + 2a_yΔy$

$v_{fy} = \sqrt{2a_yΔy}$

Δy = max height to the ground (+)
$a_y$ = gravity (+)

$v = v_0 + at$

$v_{fy} = v_{0y} + a_y t_2$

$v_{fy} = 0 + a_y t_2$

$t_2 = \frac{v_{fy}} {a_y}$

Final Distance

$v_{0x}= v_{flight} \cos{45°}$

$Δx = v_{0x} t + 1/2 a t^2$

$Δx = v_{0x} t + 0$

$Δx = v_{0x} (t_1 + t_2)$

Last edited: Mar 8, 2014
2. Mar 8, 2014

### haruspex

That all looks right. Do you have a question?

3. Mar 8, 2014

### layout4it

What should I plug in for the $Δx_0$ and $Δx_f$ for $PE_{spring}$?

4. Mar 8, 2014

### haruspex

It's not entirely clear, but I think the idea is that the spring starts off compressed by .0011+.0088 and finishes compressed by .0011. I.e. there is some end stop preventing it being compressed by anything less than .0011.

5. Mar 9, 2014

### layout4it

When I plug in the data I get the final distance in the x direction to be about .4m which seems unreasonable.