Box, mass m1, at rest against a compressed spring, spring constant k and compres

Question

Box, mass m1, at rest against a compressed spring, spring constant k and compression of ?x, at the top of a ramp, height h, is released. It slides down the ramp, angle ?, with friction, coefficient ?. Half way down the ramp, it bumps into and sticks to another box, mass m2, and together, they slide to the bottom. If m1 = 2.0 kg, m2 = 3.0 kg, k = 500 N/m, ?x = 0.40 m, h = 5.0 m, ? = 0.30, ? = 30.0O, how long does it take both boxes to reach the bottom (starting from when the first box is released)? The spring is parallel to the ramp surface and the ?x is measured parallel to the ramp surface.

Explanation / Answer

divide this probem into 3 parts :

a) from start to equilibrium point of spring

b) between equilibrium point and half ramp height

c) between half ramp height and bottom

part A)

spring energy = Es = 1/2kdelx^2

delx = 0.40 , k = 500N/m

Es = 1/2*500*(0.40)^2

Es = 40 J

the friction works

Wf = umgcos30*delx

Wf = 0.3*2*9.81*cos30*0.4 = 2.039 J

potential energy

U = mgh

U = 2*9.81*0.4sin30

U = 3.924J

total energy at equllibrium

= 40 + 3.924 - 2.039 = 41.885 J

= 1/2mv^2

41.885 = 1/2 mv^2
v = sqrt(2*41.885/2) = 6.472 m/s
find acceleration:
v^2 = 2as
a = v^2/(2s)
a = 6.472^2/(2*0.4) = 52.358 m/s^2
find time for this section
v = at
6.472/52.358 = t
t = 0.1236 s

part B)

Fg = mgsin30 = 2*9.81*0.5 = 9.81 N
friction force = on m1 = ?mgcos30 = 0.3*2*9.81*cos30 = 5.097 N
net force on m1: 9.81 - 5.097 = 4.7125 N
acceleration a = Fnet/m = 4.7125/2 = 2.356 m/s^2
v at half ramp height:
v^2 = v0^2 + 2as with s = (2.5 - 0.4*sin30)/sin30
v^2 = 6.472^2 + 2*2.356*(2.5 - 0.4*sin30)/sin30)
v = 7.9725 m/s
time for this section:
v = v0 + at
t = (v - v0)/a = (7.9725 - 6.472)/2.356
t = 0.6369 s

Part C)

velocity after impact with m2:
v = (m1v1)/(m1+m2) = 2*7.97/5 = 3.19 m/s
gravitational force on (m1+m2):
Fg = mgsin30 = 5*9.81*0.5 = 24.525 N
friction force on (m1+m2)
Ff = ?mgcos30 = 0.3*5*9.81*cos30 = 12.743 N
net force on (m1+m2) is
Fnet = 24.525 - 12.743 N = 11.7814 N
acceleration = Fnet/m = 11.7814/5 = 2.356 m/s^2
velocity at end of ramp is
v^2 = v0^2 + 2as
v^2 = 3.19^2 + 2*2.356*2.5/sin30
v = 5.808 m/s
time for this section:
v = v0 + at
t = (v - v0)/a = (5.808 - 3.19)/2.356 = 1.111 s

total time needed = Part A time + Part B time + Part C time

0.1236 + 0.6369 + 1.111 = 1.878 s

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