ENERGY
Q: A cart moving at 5m/s collides with a spring. At the instant the cart is motionless, what is the largest amount that the spring could be compressed? Assume no friction.
a) Define the system with the energy flow diagram, then complete the energy bar graphs qualitatively.
a) Define the system with the energy flow diagram, then complete the energy bar graphs qualitatively.
b) Quantitative Energy Conservation Equation: In position A, kinetic energy is present and there is no potential energy and no gravitational energy (no height). The equation for kinetic energy is K=1/2mv^2. The m and the v are given to us in the problem. 100J= (1/2)(8.0kg)(5.0m/s)
The kinetic energy is 100J and due to the conservation of energy principle, energy cannot be created or destroyed and so the total energy in J for potential energy is 100J as well. (The total energy in position A must equal the total energy in position B).
The kinetic energy is 100J and due to the conservation of energy principle, energy cannot be created or destroyed and so the total energy in J for potential energy is 100J as well. (The total energy in position A must equal the total energy in position B).
c) Determine the maximum compression of the spring.
A: To determine the maximum compression, the equation 1/2(kl^2) (potential energy) is used and l is found. L is the max. compression--distance. The spring constant is given to us in the problem is 50N/M.
1/2(50)(l^2)=100J <--- as previously determined
Using algebra, 'l' can be calculated to be 2m. That is the amount the car is compressing the spring.
A: To determine the maximum compression, the equation 1/2(kl^2) (potential energy) is used and l is found. L is the max. compression--distance. The spring constant is given to us in the problem is 50N/M.
1/2(50)(l^2)=100J <--- as previously determined
Using algebra, 'l' can be calculated to be 2m. That is the amount the car is compressing the spring.
Energy Problem #2
Q: The trains on the viper are raised from 10m above ground at the loading platform to a height of 60m at the top of the hill in 45s. Assume that the train (including passengers) has a mass of 2500kg. Ignoring frictional losses, what power motor would be required to accomplish this task?
A: Using the formula Power=Work/Time, we can calculate the total energy required to move the train up by finding out the change in potential energy from the initial to final state. This change in energy is equal to the Work that is required. Therefore, the Work required is calculated as mass*gravity*height or Work=(2500kg*9.8m/s^2*10m) equals 1225000J. Now, power of the motor equals to W/T or 1225000J/45s is 27222.22 watts or 27.22 kW.
A: Using the formula Power=Work/Time, we can calculate the total energy required to move the train up by finding out the change in potential energy from the initial to final state. This change in energy is equal to the Work that is required. Therefore, the Work required is calculated as mass*gravity*height or Work=(2500kg*9.8m/s^2*10m) equals 1225000J. Now, power of the motor equals to W/T or 1225000J/45s is 27222.22 watts or 27.22 kW.