Work Done on a Block Dropped onto a Spring

[mb85]

A 330 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 3.3 N/cm (Fig. 7-42). The block becomes attached to the spring and compresses the spring 11 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

For part A. the formula i figured I use is = Mgd cos 180
= (.33kg)(-9.8)(.11m) cos 180 (which is -1)
= 0.39 J

For part B. I was using the formula = - 1/2Kx^2
however I ran into some either arithmatic problems or technique. here is my work.
= -(1/2)(330N/m)(.11m)^2
= -18.15 J
But I am thinking there is a problem since the answer is incorrect.

For part C. I used the formula = 1/2mv^2 = Wnet = Wg +Wa
Wg = mgd and Wa = mk
so ultimately after reworking the formula, it became
v = Square root (gd + k) times 2
My question is... is the 2 included under the square root? or outside?
Not exactly sure where i went wrong, because the answer is incorrect.

For part D. i used d = v times square root (m/k)
but since part C is incorrect, this answer is also wrong. (or maybe the formula is too?)


Can someone help me understand where i made my mistakes. Thanks!

 

[*melinda*]

Well, in question B. they ask for the spring force which is in units of Newtons.

The units in your answer are in joules, but you need them to be in Newtons in order for the answer to make sense.

 

[mooshasta]

Well, in question B. they ask for the spring force which is in units of Newtons.

The units in your answer are in joules, but you need them to be in Newtons in order for the answer to make sense.

melinda, they're asking for the work done by the spring force, not the force.

mb85, I think you just made an arithmetic error... I did the same calulation and got -2 J. Is this correct?

For part C... I'm not sure what you meant Ma to be. 1/2mv^2 = -Wnet = -(Wg + Ws), where Ws is the work done by the spring. Wg and Ws you found in parts A and B, respectively, so it's pretty much a matter of just inputting the values and solving for v. Pretty straightfoward, right?

 

[mb85]

yeah, they were algebra mistakes.

but i still cannot get part D, when the speed is doubled for the max. compression.
can someone please help me.

 

[mb85]

cancel that. i got it. thanks!