A 24kg box slides 4.0 m down the frictionless ramp shown in the figure(Figure 1)
ID: 1260354 • Letter: A
Question
A 24kg box slides 4.0 m down the frictionless ramp shown in the figure(Figure 1) , then collides with a spring whose spring constant is 250N/m .
What is the maximum compression of the spring?
At what compression of the spring does the box have its maximum speed?
A 24kg box slides 4.0 m down the frictionless ramp shown in the figure(Figure 1) , then collides with a spring whose spring constant is 250N/m . What is the maximum compression of the spring? At what compression of the spring does the box have its maximum speed?Explanation / Answer
we will do the first part via energy conservation
the box compresses the spring a distance x, the total distance it moves down the plane is 4+x
since the plane is inclined at 30deg, the vertical drop h the box makes is given by
sin30 = h /(4+X)
h = (4+x)sin 30
= 0.5(4+x)
this must equal the gain in elastic PE
1/2 k x^2 = m g (0.5(4+x))
1/2 k x^2 = 2 m g + m g x/2
1/2 k x^2 - m g x/2 - 2 m g=0
(0.5)(250N/m)x^2 - (24 kg)(9.8 m/s^2)x/2 - 2 ( 24 kg ) (9.8 m/s^2) = 0
125 x^2 - 117.6 x - 470 =0
x=2.465 m
2) realize that gravity will accelerate the box down the plane with an acceleration g sin30 = 4.9m/s/s
once the box hits the spring, the spring will exert a force up the plane the maximum velocity will occur when the total force on the box is zero; once the box compresses the spring enough so that the total force acts up the plane, the box will start slowing down
therefore, we want to find the value of x such that
k x = m g sin(theta)
k = mg sin30/k
= 24 kg x 9.8m/s^2 x sin 30/ (250N/m)
= 0.47m
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