Hopefully someone double checks what I am going to tell you, but here it goes:
Work=KE
W=∆KE=Kf-Ki= 1/2(0.17kg)(4.7m/s)-1/2(0.17)(3.7m/s)=1/2(0.17kg)[4.7m/s-3.7m/s]
heat=q=∆KE, which is lost due to the billboard rolling down the pool table. What is the difference between the KEintial and KEfinal?
∆E=q + W
I am not sure about what I have told you, but someone will check and hopefully give you the correct response if I am wrong.
The kinetic energy of a rolling billiard ball is given by KE=1/2mv2. Suppose a 0.17-kg billiard ball is rolling down a pool table with an initial speed of 4.7m/s . As it travels, it loses some of its energy as heat. The ball slows down to 3.7m/s and then collides straight-on with a second billiard ball of equal mass. The first billiard ball completely stops and the second one rolls away with a velocity of 3.7m/s . Assume the first billiard ball is the system
Calculate W, Calculate q, Calculate ΔE .
3 answers
Typo:
W=∆KE=Kf-Ki= 1/2(0.17kg)(3.7m/s)-1/2(0.17)(4.7m/s)=1/2(0.17kg)[3.7m/s-4.7m/s]
W=∆KE=Kf-Ki= 1/2(0.17kg)(3.7m/s)-1/2(0.17)(4.7m/s)=1/2(0.17kg)[3.7m/s-4.7m/s]
Read the question wrong
Work=KE
W=∆KE=Kf-Ki= 1/2(0.17kg)(0m/s)-1/2(0.17)(4.7m/s)=1/2(0.17kg)[0 m/s-4.7m/s]
**Note: Work will be negative once you calculate it.
heat=q=∆KE, which is lost due to the billboard rolling down the pool table. What is the difference between the KEintial and KE once it slows down.
q=∆KE=Kf-Ki= 1/2(0.17kg)(3.7m/s)-1/2(0.17)(4.7m/s)=1/2(0.17kg)[3.7m/s-4.7m/s]
**Note: q will also be negative, KE is lost meaning heat is loss
∆E=q + W
Work=KE
W=∆KE=Kf-Ki= 1/2(0.17kg)(0m/s)-1/2(0.17)(4.7m/s)=1/2(0.17kg)[0 m/s-4.7m/s]
**Note: Work will be negative once you calculate it.
heat=q=∆KE, which is lost due to the billboard rolling down the pool table. What is the difference between the KEintial and KE once it slows down.
q=∆KE=Kf-Ki= 1/2(0.17kg)(3.7m/s)-1/2(0.17)(4.7m/s)=1/2(0.17kg)[3.7m/s-4.7m/s]
**Note: q will also be negative, KE is lost meaning heat is loss
∆E=q + W
0 answers