A 0.350kg block moving vertically upward collides with a vertical spring and compresses it a distance Δy...?
A 0.350kg block moving vertically upward collides with a vertical spring and compresses it a distance Δy before coming to rest. If the spring constant is 50.0N/m, and the initial velocity is 1.1m/s, what is the distance Δy that the spring is compressed?
Answer:
Δy= 0.0462m = 4.62cm
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Answers & Comments
The initial kinetic energy of the block (i.e. at the time of collision) is
KE = mv²/2
When it comes to rest, this kinetic energy is totally transformed into
gravitational potential energy and elastic potential energy
PE(g) = mg(Δy)
PE(e) = k(Δy)²/2
The law of conservation of mechanic energy gives
mv²/2 = mg(Δy) + k(Δy)²/2, or
k(Δy)² + 2mg(Δy) - mv² = 0
Plugging in known quantities:
50(Δy)² + 2*0.35*9.8(Δy) - 0.35*(1.1)² = 0
50(Δy)² + 0.7*9.8(Δy) - 0.35*1.21 = 0
50(Δy)² + 6.86(Δy) - 0.4235 = 0
This quadratic equation has 2 roots of opposite signs.
The positive root is the solution to the problem.
Solving for positive Δy by quadratic formula:
Δy = [-6.86 + √(6.86² + 4*50*0.4235)]/(2*50)
. . .= [-6.86 + √(47.06 + 84.70)]/100
. . .= [-6.86 + √(131.76)]/100
. . .= [-6.86 + 11.48]/100
. . .= 4.62/100 meter
. . .= 4.62 cm