I have the answers already, need to know the steps to get them. An amusement par
ID: 1509084 • Letter: I
Question
I have the answers already, need to know the steps to get them.
An amusement park ride consists of a ramp that is sloped at an angles of 32.9 degrees with respect of the vertical with a srping located at the top of the ramp with a spring constant of 190 N/m. A person will sit in a sled on the ramp, and then the spring is compressed by the operator by an amount of 2 m. The spring is then released by the operator and the person plus sled are pushed down the ramp by the spring. The sled will become separated from the spring once the spring returns to rest length. Also, there is friction between the sled and the ramp which has a coefficient of kinetic friction of .7. Consider the person plus sled as one object with a mass of 51.9 kg.
After the person has traveled down the ramp a distance of 5m, the person will have a speed of 7.727 m/s, the spring has returned to rest length, and you determine that 581 J of thermal energy went into the ramp as the person travels this distance.
1. Use everything as the system.
a. What is the work done on the system? Answer: 0J
b. What is the change in KE of the system? Answer:1549J
c. What is the change in thermal energy of the system (there is friction under the spring as well, so the friction will be present for the entire displacement) Answer : 967.9J
d. What is the change in gravitational PE of the system? Answer: -2137J
e. What is the change in spring potential energy of the system? Answer: -380J
2. Use the person plus sled as the system
a. What is the change in KE of the system? Answer: 1549J
b. What is the change in thermal energy of the system? Answer: 386.9J
c. What is the total work done on the system? Answer: 1936J
d. What is the work done by gravity on the system? Answer: 2137J
e. What is the work done by the spring on the system? Answer: 380J
f. What is the work done by friction on the system? Answer: -581.1J
Explanation / Answer
1. work done on the ssytem = 0 [ as everything is the system ]
2. Change in KE of the system = 0.5*m*v^2 = 0.5*51.9*7.727^2 = 1549.3844 J
3. Change in TE of the system = f*s = mu*mg*cos(theta)*s = 0.7*51.9*9.8*cos(90-32.9)*5 = 966.942 J
4. Change in spring PE = 0 - 0.5kx^2 = - 0.5*190*4 = -380 J