Spring Constants, Work, and Force

[NyghtLyght]

Homework Statement

Suppose for some unexplained reason you are skateboarding and wanted to use a spring to launch you off a ramp. Assuming you have a mass of 70 kg, if the spring has a spring constant k = 625N/m and is compresed 5m.
1.If the ramp has a 60° and is 10m long, will you make it off the ramp?
2. How high will you go?
3. How fast will it be going when it enters the ramp (ignore friction)?

Homework Equations

1/2mv(intial)squared + mgh(initial) + 1/2k(initial)xsquared = 1/2mv(final)squared + mgh(final) +1/2k(final)xsquared

The Attempt at a Solution

To get the first answer, the second question has to be answered. I got 1/2k(intial)xsqured = mgh(final). 1/2(625)(5)squared = 70(9.8)h(final). 7812.5 = 686h(final). 11.4m=h(final).
After this, I have no clue how to get the first and third question.


Thanks ahead of time.

 

[voko]

To answer #1, you need to determine whether the result in #2 is higher than the end of the ramp.

In #3, I do not understand what "it" is.

 

[NyghtLyght]

The it in #3 is the skateboard... I think.

And the answer to #2 is 11.4m=h(finial). But I don't understand how to get the answer to number one.

 

[NyghtLyght]

I think I have the answers. Thanks for the help.

 

[Nickie]

I would like to first commend you for your attempt at solving the problem and using the appropriate equations. Your calculation for the final height of the skateboarder is correct, and it is important to note that this is the maximum height they will reach on the ramp.

To answer the first question, we need to consider the angle of the ramp. Since the ramp is at a 60° angle, the horizontal distance traveled will be 10m*cos(60°) = 5m. This means that the skateboarder will just make it off the ramp, as the maximum height they can reach is also 5m.

To answer the third question, we can use the conservation of energy equation you have written. We know the initial potential energy (1/2k(initial)x(initial)^2) and the final potential energy (1/2k(final)x(final)^2) are both zero, as the skateboarder will start and end at ground level. So we can rearrange the equation to solve for the final velocity: v(final) = sqrt((mgh(final) + 1/2k(initial)x(initial)^2)/m). Plugging in the values, we get v(final) = sqrt((70*9.8*11.4 + 1/2*625*5^2)/70) = 10.6 m/s.

However, it is important to note that this calculation does not take into account friction and air resistance, which will affect the actual velocity of the skateboarder. But for the purposes of this problem, we can assume that the skateboarder will be traveling at approximately 10.6 m/s when they enter the ramp.

I hope this helps you better understand the concepts of spring constants, work, and force in this scenario. Keep up the good work in your studies!