From Autumn 2000 Student Performance

1. (25 points) Ch. 8. A fish pulls on a fishing pole that is propped near the edge of the shore as shown. The fish pulls with a force of magnitude F = 100 N, just like that quiz problem from Ch. 8. The angle that the force makes with a line parallel to the pole is shown to be 60 degrees. The pole is 2.0 m long. The pole can rotate about the specially constructed axis indicated in the figure.

(a) ( 23 points) What is the magnitude of the torque exerted on the
pole ?

(b) (2 points) Does the pole rotate clockwise or counterclockwise ?
Solution :
(a) Torque = 100 (2 sin 60) = 173 Nm

(b) Clockwise

 

2. (30 points) Ch. 9. A mattress is partially submerged under the water. The mass of the mattress is 2.0 kg, just like that quiz problem from
Chapter 9. The mattress has top and bottom area = 2.0 m2. On top of the mattress is a block of mass M. The total thickness of the mattress is 0.200 m. The length exposed above the surface of the water is 0.080 m.
Please study the figure below before you attempt to answer the questions.

(a)(6 points) What is the length H that is under the water ? (This is NOT a trick question. It requires no physics, just simple algebra to answer the question. Look at the picture. )

(b) (6 points) What is the volume of the mattress under the water ? (This is NOT a trick question. It requires no physics, just simple algebra to answer the question. Use the definition of volume.)

(c) (9 points) What is the buoyant force FB acting upward ?

(d) (9 points) What is the mass M of the block ?

Solution:

(a) 0.12 m = 0.200 - 0.08

(b) VS= A(0.12) = 2(0.12) = 0.24 m3

(c) FB = (water density)VSg = 1000(0.24)(9.8) = 2352 N

(c) 2352 = Mg + 2g where mass of mattress = 2 kg

M = (2352 - 2g)/g = (2352 - 2*9.8)/9.8 = 238 kg

 

3. (25 points) Ch. 10. An air bubble has a volume of 1.30 cm3 when it is released by a submarine 200 m below the surface of a lake, similar to that quiz problem from Chapter 10 . What is the volume of the bubble when it reaches the surface ? Assume the temperature of the air remains constant while bubble rises.

Solution:

P1 = Patm + (water density)g h

P1 = 1.013x105 + (1000)(9.8)(200)

P1 = 2.016x106 N

P1V1/T1 = P2V2/T2
Note that the temperature cancels! Also P2 = Patm = 1.013x105

V2 = (P1/P2)*V1 = (2.016/1.013)*1.3 = 27 cm3 

Extra Credit = 20 points !

4. Ch. 11. Lead bullets, each of mass 1.00 g, are heated to 200oC. How many pellets must be added to 500 g of water that is initially at 20.0oC to make the final equilibrium temperature Tf = 25.0 oC ? Neglect any heat transfer to or from the container ( Hint: See the hint to #11 of Quiz 10 for Ch. 11.)
Solution: See Hint to Problem 11, Ch. 11 by clicking here. Answer = 467 pellets