Gravitional Potential Energy and Elastic Potentional Energy

[y201]

Homework Statement

1) A rifle shoots a spring of mass 0.008kg and with a spring constant of 350 N/m. You wish to hit a target horizontally a distance of 15 m away by pointing the rifle at 45 degrees above the horizontal. How far should you extend the spring in order to reach the target?

Homework Equations

1) d=v*t, v2=v1+a*t, d=v1(t)+(0.5)(a)(t)

The Attempt at a Solution

im lost :(

Homework Statement

A bungee cord need to transfer 2,000,000 J of energy. A 10 kg mass extends the bungee cord 1.3m. What is the maximum extension of the bungee cord?

Homework Equations

F=kx, E=(0.5)(k)(x^2) W=F*d

The Attempt at a Solution

lost again :(

 

[cupid.callin]

you need to show what you've tried in the question

 

[y201]

you need to show what you've tried in the question

sadly, i don't even know where to start.

 

[AlexChandler]

Try to start by finding the maximum height that the spring must reach. Then try to consider changes in energy.

 

[y201]

Try to start by finding the maximum height that the spring must reach. Then try to consider changes in energy.

can you start me off with few steps please?

 

[AlexChandler]

can you start me off with few steps please?

Sure. Try to combine the kinematics equations by eliminating the time in order to reach the trajectory equation. It should look like this.

y = tan \theta_0 x - \frac{g cos^2 \theta_0}{2 V_0^2} x^2

Then you should be able to find the maximum height.

 

[y201]

Sure. Try to combine the kinematics equations by eliminating the time in order to reach the trajectory equation. It should look like this.

y = tan \theta_0 x - \frac{g cos^2 \theta_0}{2 V_0^2} x^2

Then you should be able to find the maximum height.

this equation is for the second question right?

 

[AlexChandler]

this equation is for the second question right?

No it is for the spring problem.

 

[cupid.callin]

you can also do it the traditional way

find the time of flight using y = ut + .5gt₂

now as horizontal speed don't change throughout the motion: distance you need to reach, d = u_X * t

so you have u_X viz ucos45

now use energy conservation

.5 kx² = .5m^v