/10 points Previous Answers SerPSE9 7 P.063 Nous Ask Your Teacher An inclined pl
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/10 points Previous Answers SerPSE9 7 P.063 Nous Ask Your Teacher An inclined plane of angle 20.0° has a spring of force constant k 535 N m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m- 2.41 kg is placed on the plane at a distance d- 0.276 m from the spring. From this position, the block is projected downward toward the spring with speed v 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest? Your response differs from the correct answer by more than 10%. Double check your cale lations m or Need Help?Read ItExplanation / Answer
We have no information about friction, so let's assume that it is negligible.
Also assume that the angle of the slope is measured from the horizontal.
Assume ideal spring
The initial kinetic energy of the block plus the change in potential energy will equal the spring potential energy at maximum compression
KE + PE = PS
½mv² + mgh = ½kx²
mv² + 2mgh = kx²
h = (d + x)sin
mv² + 2mg(d + x)sin = kx²
0 = kx² - 2mgxsin - mv² - 2mgdsin
which is a quadratic equation in x
0 = 535x² - 2(2.41)(9.81)xsin20 - (2.41)0.750² - 2(2.41)(9.81)0.276sin20
0 = 535x² - 16.172x – 5.819
x = (16.172 ±(16.172² - 4(535)(-5.819))) / (2(535)
x = 0.12 m (ans)
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