How far does the spring compress? I have no idea...?
Im very confused with this problem please show me how to do it =)
A block of mass m = 0.70 kg slides along a horizontal table with speed v0 = 1.00 m/s. At x = 0 it hits a spring with spring constant k = 26.00 N/m and it also begins to experience a friction force. The coefficient of friction is given by μ = 0.100. How far did the spring compress when the block first momentarily comes to rest?
Hint!Use the work-kinetic energy theorem knowing that after the block hits the spring, both the elastic force and the friction force act on it.
- dapromaLv 47 years agoBest Answer
K = 1/2mv^2
K = .35J
W = Fd
F compression will be W/d
sum of the forces W/d-Ff = -kx
W/d - mg(mew) = kx (x and d are same thing)
.35 / x - .7*9.81*.1 = 26x
26x^2 + .7*9.81*.1x - .35 = 0
quadratic formula (-.6867 +/- 6)/52 do the -, to get .102 meters
- 3 years ago
enable's answer all the incorrect solutions that I even have seen right here: a million - there is gravity acting on each satellite tv for pc that orbits the Earth. 2 - there is not any longer the comparable quantity of gravity acting on our satellites (ISS lined) as there is on the exterior of the Earth. The gravity field decays because of the fact the sq. of the gap. hence, satellites journey a critically much less quantity of gravity than we do on the exterior. 3 - The astronauts do journey gravity. 4 - speed isn't a rigidity. Gravity isn't "canceled" out via their speed. 5 - A sky diver isn't weightless. there's no merchandise to withstand their "falling" action and supply a rigidity equivalent in magnitude to their "weight". They nonetheless are experiencing the gravitational rigidity. 6 - The centrifugal rigidity does not exist! It in ordinary terms exists in an accelerating physique of reference. If we are in a static physique of reference, then there's no centrifugal rigidity! in ordinary terms centripetal rigidity! So what's incredibly going on? because of the fact the OP reported, the gravitational rigidity is what's protecting the ISS besides as another satellite tv for pc in orbit, via inflicting centripetal acceleration. the gap station is moving very quickly far flung from the Earth tangentially. The centripetal acceleration reasons a transformation in speed (transformations the direction). that's why you will possibly be able to desire to velocity up or slow right down to certainly substitute elevation above the Earth. So in essence, the astronauts are continually falling returned in the direction of the Earth, yet they are moving so quickly forwards that via the time they fall, they have already moved to a clean area. hence the perpetual "unfastened-falling". Gravity continues to be very lots affecting them.
- oldprofLv 77 years ago
Upon initial contact, when deformation is dX = 0, the block as total energy TE = KE = 1/2 mU^2 and that is converted into work QE = FdX = μmg dX against friction and stored energy PE = 1/2 kdX^2.
From the conservation of energy law, we can write:
TE = 1/2 mU^2 = μmg dX + 1/2 k dX^2 = .35*1^2 = .35 Joules.
We solve the quadratic for dX. 1/2*26*dX^2 + .1*.7*9.8 dX - .35 = 0 = 13 dX^2 + .686 dX - .35 whose solution is dX = ,139 m. ANS.
- 3 years ago
enable's answer each and all of the incorrect solutions that I even have seen right here: a million - there is gravity appearing on each satellite tv for pc that orbits the Earth. 2 - there is no longer the comparable quantity of gravity appearing on our satellites (ISS secure) as there is on the exterior of the Earth. The gravity field decays via fact the sq. of the gap. as a result, satellites adventure a severely much less quantity of gravity than we do on the exterior. 3 - The astronauts do adventure gravity. 4 - velocity isn't a tension. Gravity isn't "canceled" out by using their velocity. 5 - A sky diver isn't weightless. there is no merchandise to stand up to their "falling" action and furnish a tension equivalent in value to their "weight". They nevertheless are experiencing the gravitational tension. 6 - The centrifugal tension does not exist! It merely exists in an accelerating physique of reference. If we are in a static physique of reference, then there is no centrifugal tension! merely centripetal tension! So what's somewhat occurring? via fact the OP pronounced, the gravitational tension is what's conserving the ISS to boot as another satellite tv for pc in orbit, by using inflicting centripetal acceleration. the area station is shifting very quickly removed from the Earth tangentially. The centripetal acceleration motives a transformation in velocity (differences the course). that's why you need to velocity up or sluggish all the way down to incredibly replace elevation above the Earth. So in essence, the astronauts are continually falling back in the direction of the Earth, yet they're shifting so quickly forwards that by using the time they fall, they have already moved to a sparkling region. consequently the perpetual "unfastened-falling". Gravity continues to be very lots affecting them.
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- ReB94Lv 47 years ago
change in KE of the block = 0 - 0.35 = -0.35 J
-0.35 = net work done
work done by spring force = -0.5 * k * x^2
work done by frictional force = 0.1 * mg * x