Question

1. a crossbow is drawn back 30 cm and has a spring constant of 814 n/m. a 0.035 kg arrow is loaded into it.

a. when the arrow is fired, what will be its speed when it leaves the bow?
b. if the archer fires the arrow straight upward, what will be the maximum height of the arrow? (the archer intends to move out of the way of its return.)

1. the top thrill dragster at cedar point goes over a large hill after launching at high speed. the hill is 128 m above the launch point, and

the car has a mass of about
700 kg.
a. what speed is
required to make it
over the hill?
b. if the work done to
accelerate the car
is done over 16 m,
what force is
necessary to
accelerate the car?

1. how much work is done by a group of football players who push a sled at an angle of 15° to the horizontal with a force of 2,200 n if the sled only moves 11 m?

b. what is the power of the football players if it takes them 8 seconds to move the sled?
bonus: what is the relationship between watts and horsepower?
bonus 2: other than \conservation of energy\, what is another law of thermodynamics?

Explanation:

Problem 25a

Step1: Convert units

Convert 30 cm to 0.30 m.

Step2: Apply energy conservation

Elastic potential energy = Kinetic energy: $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$
Simplify: $v = \sqrt{\frac{kx^2}{m}}$
Substitute values: $v = \sqrt{\frac{814 \times (0.30)^2}{0.035}} = \sqrt{\frac{814 \times 0.09}{0.035}} = \sqrt{\frac{73.26}{0.035}} \approx \sqrt{2093.14} \approx 45.75$ m/s

Problem 25b

Step1: Apply energy conservation

Kinetic energy = Gravitational potential energy: $\frac{1}{2}mv^2 = mgh$
Simplify: $h = \frac{v^2}{2g}$
Substitute $v \approx 45.75$ m/s, $g = 9.8$ m/s²: $h = \frac{(45.75)^2}{2 \times 9.8} \approx \frac{2093}{19.6} \approx 106.8$ m

Problem 26a

Step1: Apply energy conservation

Kinetic energy = Gravitational potential energy: $\frac{1}{2}mv^2 = mgh$
Simplify: $v = \sqrt{2gh}$
Substitute $h = 128$ m, $g = 9.8$ m/s²: $v = \sqrt{2 \times 9.8 \times 128} = \sqrt{2508.8} \approx 50.1$ m/s

Problem 26b

Step1: Calculate work needed

Work = Gravitational potential energy: $W = mgh = 700 \times 9.8 \times 128 = 878080$ J

Step2: Find force

Work = Force × distance: $F = \frac{W}{d} = \frac{878080}{16} = 54880$ N

Problem 27

Step1: Calculate work done

Work = $Fd\cos\theta$
Substitute $F=2200$ N, $d=11$ m, $\theta=15^\circ$: $W = 2200 \times 11 \times \cos15^\circ \approx 2200 \times 11 \times 0.9659 \approx 23270$ J

Problem B

Step1: Calculate power

Power = $\frac{Work}{Time} = \frac{23270}{8} \approx 2909$ W

Bonus

Step1: Relate Watts and horsepower

1 horsepower ≈ 746 Watts

Bonus 2

Step1: Name another thermodynamics law

Second Law of Thermodynamics (e.g., entropy increases in isolated systems)

Answer:

25a: ~46 m/s
25b: ~107 m
26a: ~50 m/s
26b: 54880 N
27: ~23300 J
B: ~2910 W
Bonus: 1 horsepower ≈ 746 Watts
Bonus 2: Second Law of Thermodynamics (e.g., entropy of an isolated system always increases)