Question

1. a ball is pushed down a hill. at which point on this hill will the ball have the most kinetic energy? a. point a b. point c c. point d d. point e 12. what is the standard unit for power? a. newtons (n) b. joules (j) c. watts (w) 13. how much work is the person in the image on the right doing if the mass of the weights are 50 kg total and he lifted them 4 meters? (hint: solve for force first → look at the equations) 14. a lightbulb with a power of 60 watts is left on for 180 minutes. how much work is done during that time? 15. a car has a mass of 2,000 kg and is traveling at 28 meters per second. what is the cars kinetic energy? 16. a bullet from a policemans handgun travels at 200 meters per second and has a mass of 0.02 kg. what is the bullets kinetic energy? 17. if a man moves a large box that weighs 10 newtons 20 meters in 30 seconds, how much power was used? 18. if 68 w of power is produced in 18 seconds, how much work is done?

Explanation:

11.

Step1: Kinetic energy and height

Kinetic energy \(K = \frac{1}{2}mv^{2}\), and potential energy \(U = mgh\). As the ball rolls down, potential energy is converted to kinetic energy. The lower the height \(h\), the more potential energy is converted to kinetic energy.

Step2: Analyzing points

Point \(D\) is the lowest point on the hill. At this point, the ball has the least potential energy (since \(h\) is minimum) and the most kinetic energy (because energy is conserved \(E = K+U\) and \(U\) is minimum at \(D\)).

Power is defined as the rate of doing work \(P=\frac{W}{t}\). The standard unit of power in the SI system is the Watt (W). Newtons (N) is the unit of force (\(F = ma\)), Joules (J) is the unit of work (\(W = Fd\)) or energy (\(E\)).

Step1: Calculate force

Using \(F = mg\), where \(m = 50\space kg\) and \(g= 9.8\space m/s^{2}\). So \(F=50\times9.8 = 490\space N\)

Step2: Calculate work

Using \(W = Fd\), with \(F = 490\space N\) and \(d = 4\space m\). Then \(W=490\times4=1960\space J\)

Answer:

C. Point D

12.