Question

A spring is fixed at the bottom end of an incline of inclination 37 degrees. A small block of mass 2 kg starts slipping down the incline from a point 4.8m away from the spring. The block compresses the spring by 20 cm, stops momentarily and then rebounds through a distance 1 m up the incline. Find (a) the friction coefficient between block and plane and (b) The spring constant 'k'

Answer 1

4.8mgsin(35°) - 4.8μmgsin(35°) = (1/2)(0.2)^2k + 0.2μmgsin(35°)
(1/2)(0.2)^2k - 0.2μmgsin(35°) = mgsin(35°) + μmgsin(35°)
4.8mgsin(35°) - 5μmgsin(35°) = (1/2)(0.2)^2k
(1/2)(0.2)^2k - 1.2μmgsin(35°) = mgsin(35°)
mgsin(35°)(4.8 - 5μ) = (1/2)(0.2)^2k
(1/2)(0.2)^2k = mgsin(35°)(1 + 1.2μ)
4.8 - 5μ = 1 + 1.2μ
6.2μ = 3.8
μ = 0.6129
k = 976.19 N/m

Answer 2

can u please explain

Answer 3

im not good at explanations

Answer 4

i am not being able to make anything
from the answer.Any kind of explaation will be helpful

Answer 5

ahh?

Answer 6

please!!!

Answer 7

;(

Answer 8

ok you need to apply conservation of energy here.. you should know what that is, and how to apply that.. yeah?

Answer 9

ya , i am getting confused in the second case when is rebounds.

for the first case i have made the equation :

k/25 = 60 - 40u k - spring constant and u - coefficient of ffriction

Answer 10

can you write the full initial equation?
should be 4.8mgsin(35°) - 4.8μmgsin(35°) = (1/2)(0.2)^2k + 0.2μmgsin(35°), as missMob said..

Answer 11

for the first case i mean. does that match?

Answer 12

thats wat i did

Answer 13

lol. you couldnt explain it though. but yeah. well done. :)

Answer 14

thx

Answer 15

no the total distance it will travel will be 5 mtr
and height will be 3 mtrs(using sin37)
so spring energy = mgh - w.d by friction
k/25 = 2X10X3 - 2*!0*cos37 u *5

Answer 16

we can get 5 mtr as 4.8 + 20cm = 5 mte

Answer 17

yup

Answer 18

sorry i couldnt explain im really bad at explinations lol :/

Answer 19

well i guess cons of energy should be used here. i kinda understand what you did there, but i'll still stick with cons of energy.

Answer 20

i used that only:
innital energy = mgh + energy lost due to friction
final energy = 1/2kx^2

Answer 21

forgot friction?

Answer 22

no

Answer 23

oh i see it now.. the first equation looks right.
for the rebound, initial energy of the spring is 1/2kx^2.. final energy should be 1.02*(mu)*mgcos35 + mgsin(35)..

Answer 24

.02*mu*mgcos35 for the part when it rebounds to its equilibrium position.. and mu*mg*mgcos35 for the part where it goes up another 1m up the incline..

Answer 25

yeah?

Answer 26

ya i think its right

Answer 27

yeah. now two variables, two equations.. solve them to get your answer.