Physics 225, Dr. Shapiro, Spring 1998, Exam #2

Physics 225 - Dr. Shapiro

Midterm Exam #2, Spring 1998

Physics 225 – S ’98 - Dr. Shapiro Exam #2 April 9, 1998 This is a closed book exam. You may use a calculator, your 3"×5" card, and the handout. Please write clearly. Partial credit will be given.

Problem #1) A 1500 kg automobile is travelling along a straight, level road at 100 km/hr. The driver sees that a tree has fallen across the roadway, and she jams on the brakes putting the car in a skid. (a) How much kinetic energy did the car have just before the driver applied the brakes? (b) Assuming that the coefficient of kinetic friction between the tires and the road is mu_k= 0.80, how far does the car travel before it stops?

The kinetic energy is (1/2)mv² = 750 kg × (100,000/3,600 m/s)² = 5.79 × 10⁵ J.
The frictional force is (mu_k)N = (mu_k)mg = 0.8 × 1500kg × 9.8m/s² = 11,760 Nt. The work done by this force must dissipate all the kinetic energy, so Fd = 5.79 × 10⁵ J and d = 5.79×10⁵ J/11,760 Nt = 49.2 m.

Problem #2) A stone with mass of 0.5 kg is tied to the end of a 0.7 m string, and is whirled around in a vertical circle at a constant tangential speed of 3.4 m/s. What is the tension in the string (a) at the top of the circle, and (b) at the bottom of the circle?

At the top of the circle the tension force in the string, T, is directed downward, and so is the gravitational force mg. These together provide the necessary centripetal force to keep the stone moving in a circle, so T + mg = mv²/r or T = mv²/r -– mg = 0.5kg[(3.4m/s)²/0.7m -– 9.8 m/s²] = 3.36 Nt.
At the bottom the situation is reversed. The tension is upward and the gravitational force is downward, so at the bottom of the trajectory T – mg = mv²/r. Thus, T = mv²/r + mg = 0.5kg[(3.4m/s)²/0.7m + 9.8 m/s²] = 13.2 Nt.

Problem #3) A 2.0 kg block is placed at one end of a board that is 1.0 m long. The coefficient of static friction between the block and the board is mu_s = 0.30, while the coefficient of kinetic friction is mu_k = 0.20. The end of the board where the block is located is lifted until the block starts to slide. (a) At what angle does the block begin to slide? (b) How much work is done by the frictional force as the block slides down the board? (c) How much kinetic energy does the block have when it reaches the halfway point.

(Suggestion: If you are unable to determine the angle at which the block begins to slide, use the symbol "theta" for it and try to solve parts b and c in terms of "theta" .)

The block starts to slide when the downhill component of the gravitational force (mg sin(theta) ) just equals the uphill frictional force ((mu_s) mg cos(theta) ), or when mu_s = tan(theta) . Thus, theta = tan^-1(mu_s) = 16.7°.
Once the block starts to slide the frictional force is (mu_k)mg cos(theta) = 0.2 × 2.0kg × 9.8m/s² × 0.958 = 3.76 Nt, and the work done by this force = -Fd = -3.76 J (taking positive forces downhill).
The net force on the block, once it is sliding, is mg × sin(theta) - (mu_k)mg cos(theta) = 2kg × 9.8m/s²[sin(16.7°) – 0.2 cos(16.7°)] = 19.6(0.287 – 0.192) = 1.86 Nt. So from the work-energy theorem, the gain in kinetic energy is 1.86 Nt × 0.5 m = 0.931 J. This also could be solved by noting that the decrease in gravitational potential energy is mg sin(16.7°) × 0.5 m = 2.816 J, while the energy dissipated by friction is 3.76 J/2 = 1.88 J. The difference between these is just 0.93 J.

Problem #4) A marble is dropped from the top of a skyscraper. Is the work done by the Earth’s gravitational force on the marble equal to, smaller than, or larger than that done by the marble’s gravitational force on Earth? Explain.

From Newton’s Third Law we know that the force on the marble due to the earth’s gravitational field is equal in magnitude to the force on the earth due to marble’s gravitational field. However, work is

. In this case the marble drops through a significant distance, but earth does not move by any significant amount as a result of the gravitational force from the marble. The the work done by the earth’s gravitational force is larger than the work done by the marble’s gravitational force.

Problem #5) Suppose you'’re on a rooftop and you throw a ball downward to the ground below with an initial speed v. You then throw an identical ball upward with the same initial speed v. After rising, this ball also falls to the ground. At the moment each strikes the ground, how do the speeds of the balls compare? Use conservation of mechanical energy in your reasoning.

The ball thrown downward will have a larger velocity when it hits the ground. The ball thrown downward starts with a kinetic energy 0.5mv² and it gains an amount of kinetic energy equal to mgh, where h is the height it is thrown from. On the other hand, the ball thrown upward starts with 0.5mv², and since it ends up at the same height it started at it gains no additional kinetic energy.

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