Physics question that has me stumped?

Question

A 1.5-kg box that is sliding on a frictionless surface with a speed of 12 m/s approaches a horizontal spring. The spring has a spring constant of 2000 N/m. (a) How far will the spring be compressed in stopping the box? (b) How far will the spring be compressed when the box's speed is reduced to half of its initial speed?

Answer 1

f = Ma
a = dv/dt = v'
v = dx/dt = x'
a = d2x/dt2 = x''

for your spring
f = -Kx
where K = 2000
negative because force is in the opposite direction of the displacement.

and initial velocity Vi = 12, and mass M = 1.5

f = Ma ==> -Kx = Mx''
so you end up with the following differential equation
Mx'' + Kx = 0
x'' + (W^2)x = 0
natural frequency W = sqrt(K / M)

Ordinary, homogenious, no damping. Simplest second order diffeq there is...

solve the differential equation:
x will be a sinusoidal function of time t (for half a cycle).

x(t) = A * cos(W*t) + B * sin(W*t)
v(t) = W*B * cos(W*t) - W*A * sin(W*t)

boundary conditions: x(0)= 0, v(0) = Vi
B = Vi / W
A = 0

x(t) = (Vi / W) * sin(W*t)
t in range {0,pi/W}

v(t) = Vi * cos(W*t)
t in same range

(a)
at v(t0) = 0
t0 = pi / 2W

then x(t0) = Vi / W = 0.3286m

(b)
v(t1) = Vi / 2
t1 = acos(0.5) / W

x(t1) = (Vi / W) * sin(acos(0.5))
= 0.3286 * 0.866
= 0.2846m

PS. if the box became stuck to the spring on contact, then you get a sinusoidal oscillator, not just sinusoidal for a half cycle.

Answer 2

a)
For the 1.5kg box to stop when it hits the spring, it has to deccelerate from 12 m/s to 0 m/s.

Also, from the point of view of the spring, the spring will have to compress as much as the resulting displacement of the box, just after it hits the spring.

For the spring, F = -kx = -2000x N
which equals the force exerted by the box

-2000x = (1.5)(a)

Use the motion function that relates displacement, velocity and acceleration:

(v1)^2 = (v0)^2 + 2ad
(0)^2 = (12)^2 + 2ax -----> solve for a in terms of x
0 = 144 + 2ax
a = -72/x


-2000x = (1.5)(a)
-2000x = (1.5)(-72/x)
18.52x = (1/x)
x^2 = 1/18.52
x = .23m

b)
Box is reduced to half its initial speed:

(.5*12)^2 = (12)^2 + 2ax
36 = 144 + 2ax
-108 = 2ax
a = -54/x


-2000x = (1.5)(-54/x)
x^2 = 1/24.69
x = .2 m

Answer 3

the box has an initial kinetic energy of 0.5(1.5)(12^2) = 108 J
Since no energy is lost to friction, energy is conserved.
(a)
E(box) = U(spring)
E(spring) = 0.5kx^2, so 0.5(1.5)(12^2) = 0.5(2000)x^2
therefore, x = 0.329 m
(b)
E(box, initial) = E(box, 1/2 speed) + U(spring)
108 J = (0.5)(1.5)(6^2) + (0.5)(2000)(x^2)
216 = 54 + 2000x^2
162 = 2000x^2
x = 0.284 m

Answer 4

You have to use the conservation of energy for a moving object, .5mv^2, and elastic potential energy, .5kx^2, where k is the spring constant and x is the distance the spring is compressed.

a) .5mv^2 = .5kx^2 -> mv^2 = kx^2 -> ((mv^2)/k)^(.5) = x -> ((1.5*144)/2000)^.5 = .328 m

b) ((1.5*36)/2000)^.5 = .164 m

Answer 5

I'm not certain, but if this is an empty Milk Duds box, at a certain velocity it will whistle like a pig.

Answer 6

Hi i'm Mehran and I'm iranian i'm very skkiled in physics issues ok now i solve your question:
frictionless surface:E1=E2
V=12m/s
K=2000 N/M
a) (L2-L1)=? E1=E2 : K=F F=K(L2-L1)
0.5M.V.V=K(L2-L1)
0.5(1.5Kg)(12m/s)(12m/s)=2000 N/m(L2-L1)
(L2-L1)=0.054m=54mm
b)with the above formule and the way i write for you the second part of the issue is solved
0.5(1.5Kg)(6m/s)(6m/s)=2000N/m(L2-L1)
(L2-L1)=0.0135M=13.5mm
every time you have problem email me to solve that for you :(mehran_modarresi_yahoo.com)

Answer 7

For homework help, you should ask questions in
Categories --> Education & Reference --> Homework Help

Answer 8

I see two different answers up there, wonder if one is right?