8. 0.331 points | Previous Answers serpse 10 7.5 p 022.nva My Notes Ask Your Tea

Question

8. 0.331 points | Previous Answers serpse 10 7.5 p 022.nva My Notes Ask Your Teache A 1.58-kg particle is subject to a net force that varies with position as shown in the figure. The particle starts from rest at x 0. F (N) x (m 0 2 46 810 12 14 16 What is its speed at the following positions? (a) x 3.08 5.00 m m/s (b) x = 10.0 m 22.5 How would you calculate the total work done on the particle between O and 10 m? m/s (c)x-15.0 m 8.15 The force is positive throughout this entire region. Does the particle ever slow down? m/s Need Help? Watch

Explanation / Answer

From graph first we determine the work completed at the given position

a.) 15/2 J of Work completed = (1/2) mv^2

v = sqrt(15/m)=sqrt(15/1.58) =3.08 m/s

b.) 15/2 +15 J of Work completed = 1/2 mv^2

v = sqrt(45/1.58) =5.34 m/s

c.) 15 +15 J of Work completed = 1/2 mv^2

v = sqrt(60/1.58)=6.16 m/s

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