60/1009/28/2017 09:14 AM 9/28/2017 11:00 PM Print Gradebook 1 Calculator Perodic
ID: 1659078 • Letter: 6
Question
60/1009/28/2017 09:14 AM 9/28/2017 11:00 PM Print Gradebook 1 Calculator Perodic Table Question 8 of 10 Sapling Learning A block of mass m 5.78 kg is attached to a spnng which is resting on a horizontal frictionless table. The block is pushed into the compressing it by 5.00 m, and is then released from rest. The spring ns to push the block back toward the equilibrium position at x 0 m. The graph shows the component of the force (in N) exerted by the spring on the block versus the position of the block (in m) relative to equilibrium. Use the graph to answer the following questions How much work is done by the spring in pushing the block from its initial position at x=-5.00 m to x = 2.86 m? F(N) Number 14 a(m) What is the speed of the block when it reaches =2.86 m? ·1-1 -2 Number m/s What is the maximum speed of the block? -6 Number m/s Previous Check Answer 0 Next Exit HintExplanation / Answer
For a non-constant force
Work = integral over the path ( F*dx)
The equation for the line is F (x) = -6/5*x
From -5 to 2.86
Work = integral from -5 to 2.86 (-6/5*x dx)
Work = (-6/10*x^2) from -5 to 2.86
Work = (-6/10*(2.86^2 - (-5.0)^2) = 10.1 J (ans)
From work energy theorem
Work = change in energy
The energy changing is KE
Work = 1/2*m*(vf^2 - vi^2)
Assuming it starts from 0 m/s
Work = 1/2*m*vf^2
Solve for vf
Vf = sqrt (2*work / m) = sqrt (2*10.1 J / 5.78 kg) = 1.87 m/s (ans)
The max speed occurs at x = 0... that is when the force switches directions (crosses the x axis) so then the force start taking energy out. The maximum speed occurs right before the force changes direction.
So, work from -5 to 0
Work = (-6/10)*((-5.0)^2 - 0)
Work = 15 J
Into the vf equation
vf = sqrt(2*15 J / 5.78 kg) = 2.28 m/s (ans)
Hope this helps :)
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