k •Δx²/2=m1•vₒ²/2, ….. (1)
vₒ = Δx•sqrt(k/m1) =0.4•sqrt(500/2) =6.32 m/s
The distance from start to the bottom is
s=h/sinθ = 5/0.5 = 10 m.
Two parts s1 = s2 =10/2 =5 m.
The law of conservation of energy for s1:
m1•vₒ²/2 +ΔPE-W(fr) = m1•v1²/2,
Taking into account (1) we obtain
k •Δx²/2 +m1•g•h/2 - μ•m1•g•cosθ•h/2•sinθ = m1•v1²/2
v1= sqrt{ k •Δx²/m1 +g•h[1- (μ/tanθ)] }=
=sqrt{500•0.16/2 +9.8•5•[1-(0.3/tan30º]} = 7.97 m/s.
From kinematics:
a1=(v1²-vₒ²)/2s = (7.97² - 6.32²)/2•5 = 2.358 m/s²
v1=vₒ +a•t1,
t1= (v1-vₒ)/a=(7.97-6.32)/2.358 =0.6997 = 0.7 s.
The law of conservation of linear momentum
m1•v1+0 =(m1+m2) •u,
u =m1•v1/(m1+m2) = 2•7.97/5=3.2 m/s.
The law of conservation of energy for s2:
(m1+m2)•u²/2 +ΔPE-W(fr) = m1•u1²/2,
(m1+m2)•u²/2 + (m1+m2)•g•h/2 - μ•(m1+m2)•g•cosθ•h/2•sinθ = (m1+m2)•u1²/2,
u1=sqrt{u²+g•h[1-(μ/tanθ)]} =
=sqrt{3.2² +9.8•5(1-0.3/tan30º)} =5.81 m/s,
From kinematics:
a=(u1²-u²)/2•s=(5.81²-3.2²)/2•5 = 2.352 m/s²
u1=u+a•t2
t2=(u1-u)/a=(5.81-3.2)/2.352 =1.1 s.
t= t1+t2 =0.7+1.1 =1.8 s.
Box, mass m 1 , at rest against a compressed spring, spring constant k and compression of Δx, at the top of a
ramp, height h, is released. It slides down the ramp, angle θ, with friction, coefficient μ. Half way down the
ramp, it bumps into and sticks to another box, mass m 2 , and together, they slide to the bottom. If m 1 = 2.0 kg,m2= 3.0 kg, k = 500 N/m, Δx = 0.40 m, h = 5.0 m, μ= 0.30, θ= 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 Δx is measured parallel to the ramp surface.
1 answer