High schooler help with work KE theorem and springs

[toesockshoe]

Homework Statement

I attempted the problems and did them on paper. please look at the attachments below. I am a high schooler that is unsure of spring and work-kinetic theorem problems. I was especially unsure of 2 homework problems, both of which are in the attachments. can somebody tell me if the answers are correct. my teacher for some reason doesn't give us the answers (which i think is weird). If i am going wrong somewhere, please correct me. Thanks so much!

Homework Equations

KE-spring theorem

The Attempt at a Solution

please look at attachement

 

[toesockshoe]

sorry i was unable to use the format. i didnt know how to draw the pictures on the thread. sorry in advance.

 

[haruspex]

For the first problem, explain how tan() comes into it.
For the second, it doesn't help to find the speed when it hits the spring. Since it continues to descend, there is more gravitational PE lost as the spring compresses. Just compare total energy at start with that at finish.

 

[toesockshoe]

@haruspex
for the first one, would it be 10sin (35) times 0.2 for the work? then would the problem be correct?

do you mind elaborating on the second problem? i barely understand it.

 

[haruspex]

@haruspex
for the first one, would it be 10sin (35) times 0.2 for the work? then would the problem be correct?

do you mind elaborating on the second problem? i barely understand it.

Yes, sin().
In the second one, you found the speed when it hits the spring, and equated the KE at that point with the final PE stored in the spring. But as the spring compresses, the mass continues to descend, so there is more lost gravitational PE to go into spring PE.
You can do the problem in a single step by calculating the total energy when the mass is first released and equating that to the total energy when the spring is at maximum compression. To do this, you must put in an unknown x for that max compression and choose a reference height for zero gravitational PE. (It doesn't matter where you pick that height as long as you are consistent.)

 

[toesockshoe]

@haruspex hey, i fixed the second problem (i think). i basically said the potential energy when the block is released is mg(0.3+0.3) and k=0. the potential energy when the block is in the compressed spring position is -mgd+(1/2)(kd^2)... where d is the length of the compression. I let potential energy = 0 when the spring is in equilibrium and let the energy at the initial point = the energy at the final point. I got my answer to be 0.1403 meters, which is 14.03 centimeters. Can you please check this answer? I have a quiz tomorrow that I need to ace.

 

[haruspex]

the potential energy when the block is released is mg(0.3+0.3)

Why are you doubling the 0.3? Apart from that, your working looks correct.

 

[toesockshoe]

@haruspex yeah my bad. i got it. thanks! ... i drew a really messy picture and accidently counted it twice for some stupid reason.