A 2.65 kg block on a horizontal floor is attached to a horizontal spring that is

Question

A 2.65 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0380 m . The spring has force constant 810 N/m . The coefficient 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.

What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0180 m .)

Express your answer with the appropriate units.

Explanation / Answer

let the final speed of the block is v m/s

spring constant , k = 810 N/m

coefficient of friction , u = 0.42

Using work energy thoerum

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

0.5 * 2.65 * v^2 = 0.5 * 810 * (0.0380^2 - 0.0180^2) - 0.42 * 9.8 * 2.65 * 0.020

solving for v

v = 0.421 m/s

the final speed of the block is 0.421 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.