A 2.70 kg block on a horizontal floor is attached to a horizontal spring that is

Question

A 2.70 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0390 m . The spring has force constant 855 N/m . The coefficient of kinetic friction between the floor and the block is 0.37 . The block and spring are released from rest and the block slides along the floor.

Part A

What is the speed of the block when it has moved a distance of 0.0120 m from its initial position? (At this point the spring is compressed 0.0270 m .)

Express your answer with the appropriate units.

Explanation / Answer

Here ,

mass , m = 2.7 Kg

initial compression, x = 0.0390 m

spring constant , k = 855 N/m

frictional coefficinet , u = 0.37

part A)

let the speed of the block is v

Using work energy theorum

0.5 * m * v^2 = 0.5 * k * (x2^2 - x1^2) - u * mg * d

0.5 * 2.7 * v^2 = 0.5 * 855 * (0.0390^2 - 0.0270^2) - 0.37 * 2.7 * 9.8 * 0.012

solving for v

v = 0.405 m/s

the speed of the block will be 0.405 m/s

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