AP Physics Free Response Questioin Help?

Question

In my class, we are curretnling doing work/engery s tuff. I am trying to figure out how to do this problem but I dont' know how. Can someone please give me a STEP by STEP explanation of how to do this? I don't understand how to do it.

http://img228.imageshack.us/img228/4813/untitled1tj3.jpg

Answer 1

a. The spring constant for the inner spring can be determined by the fact that the force it exerts goes from 0 to 20 units (I can't read what the units are. Are they Newtons?) over a difference of 0.10 m. So its spring constant is 200 force units per meter, or 2 force units per centimeter.

b. From point A to point B, the average force exerted by the spring is 10 units. Because the force changes linearly over the distance of 0.10 m, we can use this average and calculate the work done on the 5 kilogram object as 10 x 0.10 = 1.0 Force unit-meters.

c. From point B to point C, the average force is 40 units, and the distance is 0.05 m, so the work done is 40 x 0.05 = 2.0 Force unit-meters. (Again, the force changes linearly over the 0.05 meters.

d. The initial velocity of the object will be such that its kinetic energy equals the amount of work done by the two springs to slow its velocity to zero (i.e., to absorb all of its kinetic energy).
Just use the formula for kinetic energy and set it equal to 3.0 force unit-meters.

e. To do this step, you have to recognize that the inner spring exerts a force of 20 units at point B, and (assuming its spring constant does not change) it will exert a force of 30 units at point C. So the outer spring, which provides the rest of the force, is providing 0 units at point B and 30 units at point C, a change of 30 units in 0.05 meters, or 600 units per meter (= 6 units per centimeter).

Answer 2

For the 1st one, you should hit upon the vertical distance (no longer the gap alongside the plane) between the block and the spring. The block could have kinetic capacity equivalent to m*g*h at that factor. So plug into (a million/2)mv^2 to be sure speed. For the 2d, while the block contains relax, it could have 0 kinetic capacity. considering the fact that capacity is conserved, each and all of the aptitude capacity could desire to be interior the spring. so which you will possibly decide to have an equation for PE released by capacity of the block because it falls and PE contained interior the spring because it incredibly is compressed. while they are equivalent, the block will quit. considering the fact that sin(30degrees) is a million/2, meaning that each and each cm the spring is compressed, the block falls by way of a .5cm distance. additionally, using fact the block is 6m from the splendid of the spring alongside the plane, it would be 3m from the splendid of the spring in height. utilising the splendid of the spring using fact the reference for y=0, then: E(grav) = 20kg * g * (y - 3m) E(grav) = 196 kg m /s^2 * (y - 3m) E(spring) = a million/2 ok x^2 E(spring) = a million/2 ok (2y)^2 E(spring) = 2 ok y^2 E(spring) = 400N/m y^2 so: 196kg m/s^2 * (3m - y) = 400N/m y^2 196 m(3m - y) = 4 hundred y^2 400y^2 +196m y - 588m^2 = 0 Roots are .992 (it incredibly is previous to the collision) and -a million.48m. returned, it is vertical height, so the spring could be compressed 2x that or 2.96m. permit's see if it is genuine. E(gravity) = (a million.48m + 3m) * g * 20kg E(gravity) = 878J PE(spring) = (a million/2) * ok * x^2 PE(spring) = (a million/2) * 200N/m * (2.96m)^2 PE(spring) = 876J So incredibly close to to there. the wonderful question is on the subject of the fee after the collision. the speed is at a maximum while the internet rigidity on the block has 0 vertical element. previous to the collision, gravity is pulling it downward. on the acceptable 2d of collision, the rigidity from the spring is 0, so isn't decelerating it. the speed will proceed to extend till the rigidity on the spring precisely balances the rigidity of gravity. The rigidity of gravity alongside the plane is m*g * sin30 or 98N. The spring would not attain a resisting rigidity of ninety 8 N till: F = kx x= F/ok x = 98N / 200N/m x = .49m So in basic terms after .49m of compression does the block start to decelerate.

Answer 3

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