Work Energy Theorem and Kinematics

• layout4it
In summary: Is my math wrong?In summary, the problem involves a small steel ball being launched into the air from a plunger attached to a spring. Using kinematic and work/energy equations, the final distance of the ball can be calculated by considering the launch, end of launch to maximum height, and maximum height to ground phases. The final distance can be determined by plugging in the initial and final spring compression distances into the potential energy of the spring equation. However, the result of approximately 0.4m may seem unreasonable and may require further examination.
layout4it

Homework Statement

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 into 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?

Homework 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##

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:
That all looks right. Do you have a question?

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

layout4it said:
What should I plug in for the ##Δx_0## and ##Δx_f## for ##PE_{spring}##?
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.

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

1. What is the Work Energy Theorem?

The Work Energy Theorem is a fundamental concept in physics that states the work done on an object is equal to the change in its kinetic energy. This means that work and energy are directly related and can be used to determine the motion of an object.

2. How is work calculated in the Work Energy Theorem?

Work is calculated by multiplying the force applied to an object by the displacement in the direction of the force. Mathematically, it can be written as W = F * d.

3. What is the relationship between work and energy in the Work Energy Theorem?

The Work Energy Theorem states that the work done on an object is equal to its change in kinetic energy. This means that work and energy are directly proportional to each other. As work increases, so does the object's kinetic energy, and vice versa.

4. How does the Work Energy Theorem relate to kinematics?

Kinematics is the study of an object's motion, including its position, velocity, and acceleration. The Work Energy Theorem is related to kinematics because it can be used to determine the motion of an object by calculating the work done on it and its resulting change in kinetic energy.

5. Can the Work Energy Theorem be applied to all types of motion?

Yes, the Work Energy Theorem can be applied to all types of motion, including linear, rotational, and general motion. It is a universal principle that can be used to understand and analyze the motion of any object, as long as there is a force acting on it.

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