Potential Energy of a System Problems and Solutions

Problem#1

A 1 000-kg roller coaster train is initially at the top of a rise, at point A. It then moves 135 ft, at an angle of 40.0° below the horizontal, to a lower point B. (a) Choose point B to be the zero level for gravitational potential energy. Find the potential energy of the roller coaster–Earth system at points A and B, and the change in potential energy as the coaster moves. (b) Repeat part (a), setting the zero reference level at point A.

Answer:

Given: m = 1000 kg, 1 foot = 0.3048 m, then 135 ft = 41.1 m

(a) With our choice for the zero level for potential energy when the car is at point B, UB = 0

When the car is at point A, the potential energy of the car-Earth system is given by

UA = mgy

where y is the vertical height above zero level. With 41.1 m , this height is found as:

y = (41.1 m) sin40.00 = 26.4 m

 

Thus,

UA = (1000 kg)(9.80 m/s2)(26.4 m) = 2.59 x 105 J

The change in potential energy as the car moves from A to B is

UB – UA = 0 – 2.59 x 105 J = –2.59 x 105 J

(b) With our choice of the zero level when the car is at point A, we have UA = 0 . The potential energy when the car is at point B is given by UB = mgy where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5 m. Because this distance is now below the zero reference level, it is a negative number.

Thus, 

UB = (1000 kg)(9.80 m/s2)(–26.4 m) = –2.59 x 105 J

The change in potential energy when the car moves from A to B is

UB – U­A = –2.59 x 105 J

 

Problem#2

A 400-N child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy of the child–Earth system relative to the child’s lowest position when (a) the ropes are horizontal, (b) the ropes make a 30.0° angle with the vertical, and (c) the child is at the bottom of the circular arc.

 

Answer:

(a) We take the zero configuration of system potential energy with the child at the lowest point of the arc. When the string is held horizontal initially, the initial position is 2.00 m above the zero level. Thus,

Ug = mgy = (400 N)(2.00 m) = 800 J

(b) From the sketch, we see that at an angle of 30.0° the child is at a vertical height of (2.00m)(1 – cos30.00) above the lowest point of the arc. Thus,

Ug = mgy = (400 N)(2.00m)(1 – cos30.00)= 107 J

(c) The zero level has been selected at the lowest point of the arc. Therefore, Ug = 0 at this location.

 

Problem#3

A person with a remote mountain cabin plans to install her own hydroelectric plant. A nearby stream is 3.00 m wide and 0.500 m deep. Water flows at 1.20 m/s over the brink of a waterfall 5.00 m high. The manufacturer promises only 25.0% efficiency in converting the potential energy of the water–Earth system into electric energy. Find the power she can generate. (Large-scale hydroelectric plants,

with a much larger drop, are more efficient.)

Answer:

The volume flow rate is the volume of water going over the falls each second:

V/s = (3.00 m)(0.500 m)(1.20 m/s) = 1.80 m3/s

The mass flow rate is

m/t = ρV/t = (1000 kg/m2)(1.80 m3/s) = 1800 kg/s

If the stream has uniform width and depth, the speed of the water below the falls is the same as the speed above the falls. Then no kinetic energy, but only gravitational energy is available for conversion into internal and electric energy.

The input power is

Pin = energy/t = mgy/t

Pin = (m/t)gy = (1800 kg/s)(9.80 m/s2)(5.00 m) = 88.2 kW

The output power is

Pout = (efficiency)Pin = (25%)(88.2 kW) = 22.0 kW

The efficiency of electric generation at Hoover Dam is about 85%, with a head of water (vertical

drop) of 174 m. Intensive research is underway to improve the efficiency of low head generators. 

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