I'm having trouble with a few problems involving kinetic and potential energy.
1) A .30 kg mass attached to a spring is pulled back horizontally across a table so that the potential energy of the system is increased from zero to 120 J. Ignoring friction, what is the kinetic energy of the system after the mass is released and has moved to a point where the potential energy has decreased to 70J?
2) A sled and rider with a combined mass of 50kg are the top of a hill a height of 14m above the level ground. The sled is given a push providing an initial kinetic energy at the top of the hill of 1700 J.
a) After the push, what is the total mechanical energy of the sled and rider at the top of the hill?
B) If friction can be ignored, what will be the kinetic energy of the sled and rider at the bottom of the hill?
I know that the potential energy at the top of the hill is 6860J.
Answers & Comments
Verified answer
1) 120 J
2)
a) mgh + 1700 = 8560 J
b) KEb = 8560 J
1) if there is no friction then the total mechanical energy (KE +PE) is constant and equal to the initial PE. so when the potential is reduced to 70 from 120 obviously the difference must have gone to KE. so it should be 50J
2) the total energy initially is 6860 J
so after the push the ME should br 6860 J
now at the bottom of the hill all the energy is expended on KE so at the bottom KE is 6860 J
skill skill at 0.8m is Wp=9.8m/s^2 * 1kg * 0.8m = 7.84J that's additionally finished skill as a results of fact kinetic skill at this ingredient is 0! while skill skill is comparable to kinetic skill the aptitude skill is 0.5 the finished skill as a results of fact finished skill = skill + kinetic. So while kinetic is comparable to skill that's Wp/2=3.29J i think of you comprehend kinetic skill is Wk=(m*v^2)/2 so merely exhibit v from right here and because Wk=Wp/2 it is peace of cake. v=sqrt(Wp)=2.8m/s