A 2.50kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0350m . The spring has force constant 895N/m . The coefficient of kinetic friction between the floor and the block is 0.45 . The block and spring are released from rest and the block slides along the floor.

What is the speed of the block when it has moved a distance of 0.0150m from its initial position? (At this point the spring is compressed 0.0200m .)

0.617m/s = 61.7cm/s

Explanation:

To find the values of the velocity you take into account the following formula:

 (1)

That is, total work equals the change in the kinetic energy.

v_o: initial velocity = 0

m: mass = 2.50kg

The total work is:

(2)

x: the distance in wich the spring has been compressed

d: distance of the work

you will calculate v for a distance d = 0.0150m. By replacing in (2) and taking into account (2) you obtain:

hence, the velocity is 0.617m/s = 61.7cm/s

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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