In: Physics
Consider the following four blocks, all of which are attached to
identical springs:
block A: mass = 200·g, spring stretched 10·cm
block B: mass = 800·g, spring stretched 20·cm
block C: mass = 400·g, spring stretched 20·cm
block D: mass = 400·g, spring stretched 20·cm.
(a) All the blocks are placed on a level surface that has no
significant friction. For each of the following ranking tasks
use the symbols > and = to indicate your ranking (ties are
possible), listing any that are equal in alphabetical order.
For example, if Y and Z are the same and are both greater
than X, then you would enter Y=Z>X (note: Z=Y>X would be
incorrect, since Y and Z, being equal, must be in alphabetical
order). Rank the blocks, from largest to smallest, based on the
potential energy stored in the spring attached to each block.
largest PE smallest PE
(b) The springs are released and the blocks begin to move. Rank the
blocks, from largest to smallest, based on their kinetic energy
when they reach the mid-point of their oscillation (i.e., when
their springs are unstretched)?
largest KE smallest KE
(c) Compare the relative speed of blocks A and B when they
reach their mid-points. Choose one answer only.
speed of A > speed of Bspeed of A = speed of B speed of B > speed of A
(d) Compare the relative speed of blocks A and D when they
reach their mid-points. Choose one answer only.
speed of A > speed of Dspeed of A = speed of D speed of D > speed of A
Given
block A: mass = 200·g, spring stretched 10·cm
block B: mass = 800·g, spring stretched 20·cm
block C: mass = 400·g, spring stretched 20·cm
block D: mass = 400·g, spring stretched 20·cm.
a)
elastic potential energy is U = 0.5*k*x^2
all are of same k value so U depends on x value
from the given data B = C = D > A
b)
the largest kinetic energy will be at mean position
by conservation of energy 0.5*k*x^2 = 0.5 *m*v^2
v = sqrt(k/m) (x)
k is constant ==> v = sqrt(1/m)(x) ==> k1 = 0.5*m1*(1/m1)x^2 = 0.5*x^2
means the kinetic energy depends on the stretch in the spring
so the ranking is B = C = D > A
c) speed of block A when it reach the mid point
VA = sqrt(1/m)(x) = sqrt(1/0.2)(0.1) m/s = 0.22361 m/s
VB = sqrt(1/m)(x) = sqrt(1/0.8)(0.2) m/s = 0.22361 m/s
both having same speed when they reach the midpoint
speed of A = speed of B
d)
speed of block A when it reach the mid point
VA = sqrt(1/m)(x) = sqrt(1/0.2)(0.1) m/s = 0.22361 m/s
speed of block D when it reach the mid point
VD = sqrt(1/m)(x) = sqrt(1/0.4)(0.2) m/s = 0.316 m/s
speed of D > speed of A