(0%) Problem 9: A block of mass m-a43 kg is set against a spring with a spring c

Question

(0%) Problem 9: A block of mass m-a43 kg is set against a spring with a spring constant of ki 509 Nm which has been compressed by a distance of 0.1 m Some distance in front of it, along a fiictionless surface, is another spring with a spring constant of k2-413 Nm. Otheexpertta com 33% Part (a) How far do, in meters, will the second spring compress when the block runs into it? Grade Summary 100% sin0 Submissions Attempts remaining 10 per attempt) 1 2 3 atan0 Lacotan0 sinh0 Degrees O Radians I give up! Hints: deduction per hint. Hints remaining: Feedback: deduction per feedback 33% Part How fast v, in meters per second, will the block be moving when it strikes the second spring? 33% Part (e) Now assume friction is present on the surface in between the ends ofthe springs at their equilibrium lengths, and the coefficient of kinetic friction is uk- 0.5. If the distance between the springs is x-1 m, how far dy, in meters, will the second spring now compress?

Explanation / Answer

a)since there is no friction, the conservation of energy applies.

From the conservation of energy:

1/2 k1 d1^2 = 1/2 k2 d2^2

d2 = d1 sqrt (k1/k2)

d2 = 0.1 sqrt (509/443) = 0.107 m

Hence, d2 = 0.107 m

b)From conservation of energy

1/2 k1 d1^2 = 1/2 m v^2

v = d1 sqrt (k1/m)

v = 0.1 sqrt (509/0.43) = 3.44 m/s

Hence, v = 3.44 m/s

c)The work done by the frictional force is:

Wf = u m g x = 0.5 x 0.43 x 9.8 x 1 = 2.11 J

from conservation of energy

PE1 - Wf = PE2

1/2 k2 d2^2 = 1/2 k1 d1^2 - 2.11

1/2 k2 d2^2 = 0.5 x 509 x 0.1^2 - 2.11 = 0.435

d2 = sqrt (2 x 0.435/443 = 0.044 m

Hence, d2 = 0.044 m

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