Need Help? Read C 0/1 points I Previous Answers SerPSET9 7.P063 495 N/m fastened

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Need Help? Read C 0/1 points I Previous Answers SerPSET9 7.P063 495 N/m fastened securely at the bottom so that the An inclined plane of angle 8 20.0 has a spring of force constant k kg is placed on the plane at a distance spring is parallel to the surface as shown in the figure below. A block of mass m 2.57 speed v s 0.750 m/s. d- 0.288 m from the spring. From this position, the block is projected downward toward the spring with By what distance is the spring compressed when the block momentarily comes to rest? m 063942 X by more than 10%. Double check your calculations. Your response differs from the correct answer Need Help? Read Submit Assignment Save Assignment Progress Home My Assignments nin7-2017 Advanced Instructional systems, Inc. Au rights reserved.

Explanation / Answer

Given values :

Mass = 2.57kg, K = 495 N/m, d = 0.288 m, v = 0.750 m/s and theta = 20o

Coefficient of friction = 1 (assumed for fullest dynamic friction as it is not given)

Solution:

In the flat surface, the normal force is equal to the weight(mg) of the object in motion. How ever in an inclined plane the normal force is mgcos.

Using work - energy relation,

T1 + U = T2 the final velocity is 0, The final velocity is when the block comes to rest.

Therefore (mv^2)/2 + U = 0,

0.7228+U = 0 therefore U = 0.7228 J

Using the value, the compression length can be found:

0.7228 = 0.5(K)(square of final compression length - square of initial compression length)

0.7228 = 0.5(495)(x^2 - 0)

0.7228 = 247.5(x^2)

0.00292 = x^2

x = squareroot(0.00292)

x = 0.054m

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