3. A 2.70-kg block contacts a horizontal spring at its equilibrium position (x 0

Question

3. A 2.70-kg block contacts a horizontal spring at its equilibrium position (x 0). The block is moved a distance of 0.115 m to the left, compressing the spring as shown in the figure, and is then released from rest to slide across a horizontal surface. If the spring's force constant is k 4.60 × 102 Nm and the coefficient of kinetic friction between the block and the surface is ?,- 0.255, determine the final position to which the block slides before coming to rest. (Assume the block is not attached to the spring.) Page 1

Explanation / Answer

let d is the distance travelled by the block before coming to rest.

Workdone by friction = change in mechanical energy

fk*d*cos(180) = 0 - (1/2)*k*x^2

-fk*d = -(1/2)*k*x^2

mue_k*m*g*d = (1/2)*k*x^2

d = k*x^2/(2*mue_k*m*g)

= 4.6*10^2*0.115^2/(2*0.255*2.7*9.8)

= 0.451 m

so, at the block stps at x = d - 0.115

= 0.451 - 0.115

= 0.336 m <<<<<<<<--------------Answer

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