More Physics Fun
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From: St. Peter
More Physics Fun
These are actual physic's problems, they can be answered
A level cirular track has a radius of 1600 meters. A 1000 kilogram car moves around the track with a speed of 97 meters per second. What coefficient of static friction between the tires and the track is required to keep the car from sliding sideways?
A toy car has a mass of 1.9 kg. What minimum speed must it have at the top of a loop (like in a rollercoaster ride) so that it will not fall? The radius of the loop is 1.3 meters.
A homogeneous solid 5.0 kg cylinder rolls, without slipping down an inclined plane which is 11 meters tall. Find the speed of the cylinder at the bottom of the plane if the cylinder starts from rest.
Yes, they all have answers.
A level cirular track has a radius of 1600 meters. A 1000 kilogram car moves around the track with a speed of 97 meters per second. What coefficient of static friction between the tires and the track is required to keep the car from sliding sideways?
A toy car has a mass of 1.9 kg. What minimum speed must it have at the top of a loop (like in a rollercoaster ride) so that it will not fall? The radius of the loop is 1.3 meters.
A homogeneous solid 5.0 kg cylinder rolls, without slipping down an inclined plane which is 11 meters tall. Find the speed of the cylinder at the bottom of the plane if the cylinder starts from rest.
Yes, they all have answers.
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Postmaster
Joined: Aug 2003
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From: St. Peter
lol, actually i was just kinda more intersted to see if anyone has retained this kind of knowledge since high school. Or if its one of those things that people learn then just forget. So far it seems to be one of those things people forget. Lol.
Postmaster
Joined: Jan 2005
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From: Peoria, IL
I don't remember the formula's for the relationship vetween velocity and angular acceleration so I can't do the first two. The third one is pretty easy though.
PE = mgh -> 5 x 9.8 x 11 = 539
KE = PE
KE - 1/2 mV^2 1/2 x 5 x V^2 = 539
V = 14.7meters/sec
PE = mgh -> 5 x 9.8 x 11 = 539
KE = PE
KE - 1/2 mV^2 1/2 x 5 x V^2 = 539
V = 14.7meters/sec
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Try this one. I took physics, but it's been a while.
What is the stopping distance of 7000# Super Duty on snow covered road from 60 MPH? What is the stopping distance of a 8000# Super Duty with the same coefficient of friction surface(snowy road) from the same speed? Tires brakes etc held constant. Just pick a coefficient of friction close to snow.
What is the stopping distance of 7000# Super Duty on snow covered road from 60 MPH? What is the stopping distance of a 8000# Super Duty with the same coefficient of friction surface(snowy road) from the same speed? Tires brakes etc held constant. Just pick a coefficient of friction close to snow.
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