Question

A 2.00-kg package is released on a 53.1∘ incline, 4.00 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline (Figure 1) . The coefficients of friction between the package and the incline are μs=0.40 and μk=0.20. The mass of the spring is negligible.

A. What is the speed of the package just before it reaches the spring?

B. What is the maximum compression of the spring?

C. The package rebounds back up the incline. How close does it get to its initial position?

Answer 1

m =2 kg, = 53.1⁰ , l =4 m, k =120 N/m ,μs=0.40 , μk=0.20

Part A:

Work done by nonconservative force ( friciton force) equal to change in mechanical energy

Wnc = (Kf+Uf) -(Ki+Ui)

-F_f l = [(1/2)mvf² + mgy_f ] -[(1/2)mvi² + mgy_i ] ... (1)

F_f =μk N =μkmgcos

yf = 0, yi = l sin, vi =0 vf =v

From above values equation (1) becomes

v = [2gl (sin -μk cos) ]^1/2

v = [(2*9.8*4) {(sin(53.1) - 0.4*cos(53.1)} ]^1/2

v = 7.3 m/s

Part B:

To find the maximum compression of the spring, we will use the change in kinetic energy, change in potential energy and the work done by friction. Starting from the top of the incline (the initial and final kinetic energy of the block is zero):

K +U =E

E = - Ffs

mg (yf -yo)+ [(1/2) k(sf^2 -so^2) ]= -Ff s

yf -y0 = - (l+d sin )

s = l+d, sf =d, so =0

(1/2)kd² - d (mgsin -μk mgcos) + (μk mglcos-mgl sin) = 0

[(1/2)*120*d²] - [d {2*9.8*sin(53.1)} -{0.2*2*9.8 cos(53.1)} ] + [ 0.2*2*9.8cos(53.1) - 2*9.8*4 * sin(53.1)] =0

By soliving above equation we get

d = 1.06 m

Part C:

Starting with the spring at its maximum compression, let’s find out how high the block rebounds up the incline (final and initial kinetic energy is zero).

mg (yf -yo)+ [(1/2) k(sf^2 -so^2) ]= -Ff s

yf -y0 = - (l' +d sin )

s = l' +d, so =d, sf =0

(1/2)kd² -mg(d+ l' sin) = -μkmgcos(d+l' )

l' = {kd² / [2 (mgsin +μkmgcos ) ] }- d

l ' = { 120*1.06*1.06/ [ 2(2*9.8 cos(53.1))+ (0.2*2*9.8*cos(53.1))]} - 1.06

l' = 2.68 m where l' is distance up the incline

4 -2.68 =1.32 m

So the block ends up 1.32 m lower (along the incline) than its original position