Problem 8.51 (Figure 1) depicts the potential energy of a 310-kg pallet that sli

Question

Problem 8.51

(Figure 1) depicts the potential energy of a 310-kg pallet that slides from rest down a frictionless roller conveyer ramp, then moves across a frictionless horizontal conveyer, and finally encounters a horizontal spring at x=4.0m. When the pallet encounters the spring, it momentarily comes to rest at x=6.0m.

A) What is the force constant of the spring?

B) At what height h above the horizontal conveyer did the pallet begin sliding from rest?

C) What is the angle of inclination of the conveyer ramp?

D) What will be the speed of the pallet at x=6.0m if it begins sliding from rest at x=0?

Please provide explanations along with answers. I do not understand where to begin. Thank you so much.

https://session.masteringphysics.com/problemAsset/2334515/2/fig_8-35.png

Explanation / Answer

(a) 1/2 kx2= 2250

1/2 k (7-4)2=2250

K= 500N/m

(b)the pellet comes to rest at x=6m. so all energy atx=6m is potential. so PE=1000j

mgh=1000

310(9.8)h=1000

h=0.33 m.

(c)distance along ramp=x= 2m

h= height at x=0

mgh=2500

310(9.8)h= 2500

h= 0.823

sin theta= h/x=0.823/2

theta=24.3 degrees.

(d)at x=0,PE=2500

At x= 6m, PE=1000

so KE= 2500-1000=1/2 mv2

1/2 (310)(v)2= 1500

v=3.11 m/s

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