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Course Home xMasteringPhysics: HW07 ? ? ?Secure https://session, masteringphysics.com/myct/itemView?offset-next&assignmentProblemlD-93532694; HWO7 Exercise 7.26 ? 2015 Constants PartA A 2.90 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0340 m. The spring has force constant 850 N/m What is the speed of the block when it has moved a distance of 0.0190 m from its initial position? (At this point the spring is compressed 0.0150 m.) Express your answer with the appropriate units. The coefficlent of kinetic friction between the floor and the block is 0.42 The block and spring are released from rest and the block slides along the floor Submit X Incorrect; Try Again; 5 attempts remaining Provide Feedback Next >

Explanation / Answer

Spring Constant (K) = 850 N/m

Mass of the object (M) = 2.9 Kg

Initial compression in the spring (X) = 0.0340

Suppose velocity of the object is 'V' when the block has moved by a distance of 'd" = 0.0190 m

Therefore final compression in the spring (X') = X - d = 0.0340 - 0.0190 = 0.0150 m

Coefficient of kinetic friction between the block and surface Uk = 0.42

Normal Reaction on the block by the surface = R = M x g = Mg, (Here 'g' is acceleration due to gravity = 9.8 m/sec2)

Therefor from the law of conservation of energy,

Potential Energy stored in the spring when the block is at rest initially =

Potential Energy stored in the spring when block has moved by a distance 'd' + Kinetic Energy of the block + Energy decipated against friction in moving a distance 'd' on the surface

Thus, 1/2KX2 = 1/2KX'2 + 1/2MV2 + UkRd

or 1/2KX2 = 1/2KX'2 + 1/2MV2 + UkMgd

or KX2 = KX'2 + MV2 + 2UkMgd

or 850 x (0.0340)2 = 850 x (0.150)2 + 2.9 x V2 + 2 x 0.42 x 2.9 x 9.8 x 0.0190

or V2 = 2.5724

or V = 1.6039 m/sec

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