Recent content by AnnieD

A block of mass m = 1.9 kg is dropped from height h = 77 cm onto a spring of spring constant k = 1510 N/m (Fig. 8-38). Find the maximum distance the spring is compressed. My attempt: mgh = 1/2kx^2 (1.9)(9.8)(.77) = (.5)(1510)x^2 x = 0.138m But it's not the correct answer. What am I...

Same thing. The positive value was correct, but the negative position wasn't. That's okay. :)

Nope, that didn't work either. That's okay.. I give up. Thanks for your help! :)

8J = .08Ncm? This is the new question.. well really the same question, just different values. we must apply a force of magnitude 81.0 N to hold the block stationary at x=-3.0 cm. From that position, we then slowly move the block so that our force does +7.0 J of work on the spring–block system...

I tried those answers, but they're incorrect. It's an online assignment so it tells us right away whether or not we have the right answers.

okay.. so 8 = 1/2(41)[(x)^2 - (-2)^2)] 8 = 1/2(41)(x^2 - 4) 16 = 41x^2 - 164 180 = 41x^2 x = 2.10 And for negative: 8 = 1/2(41)[(-2)^2 - (x)^2)] 8 = 1/2(41)(4 - x^2) 16 = (41)(4-x^2) 16 = 164 - 41x^2 16 - 164 = - 41x^2 -148 = - 41x^2 x = 1.90cm??

PE = mgh But there's no height.. or mass? Is the solving for F = -kx 82 = -k(-2) 82 = 2k 41 = k at least correct? I have 46min to solve this question.

No wait.. that doesn't make sense either. Why would there be more N acting on a smaller area? I'm lost!

haha that would be a good start. So does this make sense.. Convert 8J aka 8Nm to 800Ncm U = 1/2kx^2 For positive: 82 + 800 = 1/2(41)x^2 solve for x .. 6.56cm For negative: 800 - 82 = 1/2(41)x^2 solve for x.. 5.92cm ?

In Fig.7-11 http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/pict_7_11.gif we must apply a force of magnitude 82.0 N to hold the block stationary at x=-2.0 cm. From that position, we then slowly move the block so that our force does +8.0 J of work on the spring–block system; the...

Oh, okay. I get it. Didn't know that before. Thanks again! Sorry for all the questions- we have a test tomorrow, and our teacher isn't the best for explaining things fully.

Thank you! But one more question: How did you know to use F = mg(sin). I understand the force = component part. But, when I tried to draw it out.. it doesn't seem to make sense. Sin = opposite/hypotenuse .. so sin20.. then the opposite of that would be the Fg (mg). So sin20 = fg/h .. h is the...

W = F x d 583.1 = F x 5 F = 116.2N F = mg(sin20) = (35)(9.8)(sin20) = 117.3N Is this right?

Okay, so then: a) Ep = mgh = (35)(9.8)(1.7) = 583.1J W = f x d 583.1 = F x 1.7 F = 343N b) solved in a (Ep) = 583.1J c) W = F x d 583.1 = F x 5(sin20) F = 340.98N ? Still unsure about C.

These are the book's answers: a) 3.4 x 10^2N b) 5.8 x 10^2N c) 1.2 x 10^2N