1. Discussions Feb 26 Work Conservation of energy

2. Work and conservation of energy

3. Bungee jump You are contemplating doing a bungee jump. Initially you would free fall for 60 meters. The bungee cord then slows you down to your minimum height over a distance of 40 meters. What average force does the bungee cord exert on you while it is slowing you down? (assume a weight of 600 N).

4. Bungee jump You are contemplating doing a bungee jump. Initially you would free fall for 40 meters. The bungee cord then slows you down to your minimum height over a distance of 60 meters. What average force does the bungee cord exert on you while it is slowing you down? (assume a weight of 600 N). We know that the net work on you is zero: W net =W g +W b Two forces act on you: Gravity: acts over the full distance (100 m), in the direction of the distance. The work gravity does on you is: W g =F g d tot cos(  )= (600 N)(100 m)cos(0°)=+60000 J Tension: from the bungee cord, acts only over the final 40 meters of your fall. It has to do an amount of work equal, but of opposite sign, from gravity. W b =–60000 J = F b d b cos(180°) F b =-W b /d b =(60000 J)/(60 m)=1000 N (about 1.7 times your weight).

5. Roller coaster A roller coaster barely reaches the top of a hill (at very low speed) then descends 40 meters, then climbs up 25 meters. How fast is it traveling at the bottom, between the hills? How fast is it traveling at the top of the second hill?

6. Roller coaster A roller coaster descends 40 meters, then climbs up 25 meters. How fast is it traveling at the bottom, between the hills? Use conservation of energy: mgy f +½ mv f 2 =mgy 0 +½ mv 0 2 using y f =0 m, y 0 =40 m, v 0 ≈0 m/s. Mass cancels: ½ v f 2 =g(40 m) or, v f =(2g(40 m)) 1/2 =28 m/s How fast is it traveling at the top of the second hill? We can still use conservation of energy, but now with: y f =25 m, y 0 =0 m, v 0 =28 m/s. Mass still cancels: ½ v f 2 =gy 0 +½ v 0 2 –gy f v f =(v 0 2 +2g(y 0 –y f )) 1/2 =((28 m/s) 2 +2g(-25 m)) ½ =17 m/s

7. Designing a toy You are designing a child’s toy, which will consist of a spring-loaded launcher that will give a horizontal push to accelerate a model car. The spring in the launcher can be compressed up to 20 cm. The mass of the car is 0.30 kg. If the desired maximum initial speed of the car is to be no more than 5 m/s, what should the spring constant be? Hint: don’t try to calculate the work done by the spring, use conservation of energy. Is this realistic for a child? What would the maximum force needed to set the spring be? Consider that a child could weigh about 300 N.

8. Designing a toy The spring in the launcher can be compressed up to 20 cm. The mass of the car is 0.30 kg. If the desired maximum initial speed of the car is to be no more than 5 m/s, what should the spring constant be? Here, we need to use conservation of energy with the elastic potential energy: In the initial state, we start with all elastic energy (no kinetic, v 0 =0), at the release point, we have all kinetic energy (x 0 =0): ½ mv f 2 =½ kx 0 2 k=(mv f 2 )/x 0 2 =(0.3 kg)(5 m/s) 2 /(0.20 m) 2 =188 N/m Is this realistic for a child? What would the maximum force needed to set the spring be? F max =k x max = (188 N/m)(0.20 m)=40 N or about 9 lb, which seems reasonable.