Energy Begin Date: 4/6/2018 12 01.00 AM-Due Date: 4/15/2018 11 59 00 PM End Date
ID: 2038380 • Letter: E
Question
Energy Begin Date: 4/6/2018 12 01.00 AM-Due Date: 4/15/2018 11 59 00 PM End Date: 4/172018 11 59.00 PM (9%) Problem 3: Consider an object of mass ? 3kg on a frictionless table. The objed is pushed in a straight line inthe negative x direction The object experiences a repulsive force F-ax+ blx, where F is in newtons and x is the position of the object relative to the origin 33% Part (a) write an expression for the change in potential energy (U) of two objects as a function of their mital se ation (x) and their final separation () Grade Summary Potential 100% 1 32 Attempts remaining S Clte per attempt) detailed view Submit Hint 1 give up Hints:?. dedaction per hast Hatu remaining - Feedback: deduction per feedback -- 33% Part (b) If the object starts at a position ofr". 5 m apart, how much work, m joules, is required by an external force to bring it to a position of x,-3.5 m apart when a-19and b 167 ?? & 33% Part (c) lfthe object starts at rest at position x,-0. 75 m of the fixed obyect and released, at what speed v, n meters per second will the object be moving when it is at postion x-18 m apart? 44 PM ^??4/9/2018 TouchExplanation / Answer
a) 6th option is the correct answer.
U = a*(1/x2 - 1/x1) - b*ln(x2/x1)
we know, U = -integral F.dx
= -integral (a/x^2 + b/x)*dx (from x = x1 to x = x2)
= -(-a/x + b*ln(x) ) (from x = x1 to x = x2)
= (a/x - b*ln(x) ) (from x = x1 to x = x2)
= a*(1/x2 - 1/x1) - b*(ln(x2) - ln(x1))
= a*(1/x2 - 1/x1) - b*ln(x2/x1)
b) U = 19*(1/3.5 - 1/7.5) - 16*ln(3.5/7.5)
= 15.1 J
c)
delta_U = 19*(1/18 - 1/0.75) - 16*ln(18/0.75)
= -75.1 J
sogain in kinetic energy = 75.1 J
use, KE = (1/2)*m*v^2
==> v = sqrt(2*KE/m)
= sqrt(2*75.1/3)
= 7.08 m/s
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