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Two massless springs with spring constants k1=4.0N/m and k2 = 6.0 N/m are welded together at point A. A mass m = 1.0 kg that slides along the frictionless surface is incident on the springs with a velocity of V0 = .5 m/s. It hits a massless plate with velcro on the end of spring 2 and sticks.
a) In terms of k1 and k2 what is the effective spring constant keff characterizing the system, i.e. what is the spring constant of the single spring that could replace these two and give the same motion of the mass?
b)What is the maximum compression Xmax that the mass reaches from the initial equilibrium length of the springs?

The diagram has the mass sliding towards the springs where it would hit spring 2 which is joined to spring 1 at point A, with spring 1 attached to the wall.

2007-02-19 17:48:07 · 2 answers · asked by Jeff 2 in Science & Mathematics Physics

2 answers

a) k(avg)= (k1 + k2)/2
k(avg)= (4+ 6)/2=5 N/m


b) Ps=.5k(avg)X^2 (potential energy of the spring)
Ke=.5mV^2 (kinetic energy of the mass)

Pe=Ke
then
X(max)=sqrt(mV^2/k(avg))
X(max)=sqrt(1.0 (0.5)^2 / 5)
X(max)=0.316 m or 31.6 cm

2007-02-22 01:38:10 · answer #1 · answered by Edward 7 · 0 0

hi in accordance to the regulation of momentum m1v1 = m2v2 m1 = mass of bullet = 18 g v1 = speed of bullet = ninety 9 m/s m2 = mass of wood block + bullet because it is going to pierce into the wood block = 2000 + 18 = 2018g v2 = ? m1v1 = m2v2 => v2 = m1v1/m2 = 18*ninety 9/(2018) = 0.88305 m/s that's the preliminary speed of the block the suited speed of the block = vf = 0 m/s distance travelled = x m consequently, vf^2 - v1^2 = 2ax => 0^2 - 0.88305^2 = 2ax => - 0.88305^2 = 2ax ----------(a million) Now relating the rigidity by using flow with rigidity exerted with the help of the spring additionally F = -Kx the place ok = spring consistent = 680N/m and aslo F = m2a => m2a = -kx => a = -kx/m2 => a = -680x/2.018 positioned this in (a million) - 0.88305^2 = 2ax ----------(a million) => - 0.88305^2 = 2(-680x/2.018)x => - 0.77978 = - 673.93x^2 => - 0.77978/(-0.673.ninety 3) = x^2 => 0.00115706 = x^2 => x = sqrt(0.00115706) = 0.034016 m => x = 3.4016 cm Now if bullet does nott becom a factor of the wood block, then m2 = 2000 g v2 = ? m1v1 = m2v2 => v2 = m1v1/m2 = 18*ninety 9/(2000) = 0.891 m/s that's the preliminary speed of the block the suited speed of the block = vf = 0 m/s distance travelled = x m consequently, vf^2 - v1^2 = 2ax => 0^2 - 0.891^2 = 2ax => - 0.8891^2 = 2ax ----------(2) Now relating the rigidity by using flow with rigidity exerted with the help of the spring additionally F = -Kx the place ok = spring consistent = 680N/m and aslo F = m2a => m2a = -kx => a = -kx/m2 => a = -680x/2.000 positioned this in (a million) - 0.891^2 = 2ax ----------(a million) => - 0.891^2 = 2(-680x/2.000)x => - 0.793881 = - 680x^2 => - 0.793881/(- 680) = x^2 => 0.001167472 = x^2 => x = sqrt(0.001167472) = 0.0034168 m => x = 3.4168 cm so that's your desired answer. i.e., If bullet turns right into a factor of the wood container, distance = 3.4016 cm or, If bullet doesnot become a factor of the wood container, distance = 3.4168 cm with the help of how there's no longer lots distinction between the two. desire to respond to youu properly. save smiling. bye.

2016-09-29 08:52:49 · answer #2 · answered by ? 4 · 0 0

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