- #1

Elaren

- 6

- 2

- Homework Statement:
- Calculate the final speed of the block, if the force has a magnitude of 200N and the displacement is 2m along the x-direction. The system starts from rest.

- Relevant Equations:
- ##W_{nc}=\Delta KE + \Delta PE##

Note: wording is ambiguous so I assumed spring started from equilibrium, in which case it stretches as we go downslope. Final height (at lower point on slope) is 0.

Distance along slope = Distance the spring stretches = d= ##s_f## = ##2/cos{\theta}## =2.13

Height change = h = ##2 tan{\theta}## = .7

##Fd=\frac{1}{2} m v_f^2 - mgy_i + \frac{1}{2} k s_f^2##

##426=50 v_f^2 - 686 + 9074##

Leads to a square root of a negative number, so no real solution

I think this problem is wrong as written with a spring that strong. I found this same diagram being used for a perfectly reasonable oscillation problem, so I think it was taken and reused incorrectly.

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