A 2.10 kg block on a horizontal floor is attached to a horizontal spring that is

Question

A 2.10 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0370 m . The spring has force constant 900 N/m . The coefficient of kinetic friction between the floor and the block is 0.36 . The block and spring are released from rest and the block slides along the floor.You are asked to design a spring that will give a 1110 kg satellite a speed of 2.30 m/s relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00g.

What is the speed of the block when it has moved a distance of 0.0110 m from its initial position? (At this point the spring is compressed 0.0260 m .)

This is how I went about the problem.

Explanation / Answer

Initially,

Spring potential energy = kx^2 /2 = 900 x 0.037^2 /2 = 0.616 J

KE = 0

finally ,

Spring PE = 900 x 0.0260^2 /2 = 0.3042 J

KE = mv^2 /2

from initial to final position work done by friction = -f.d

= -u mg d = -0.36 x 2.10 x 9.81 x 0.0110

= - 0.0816 J

Using eenrgy conservation,

final - initial = work done by friction

(0.3042 + mv^2 /2 ) - ( 0.616 + 0 )   = - 0.0816

2.10v^2 /2 = 0.2302

v = 0.468 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.