EGET 3171


EGET 3171

Final Review Practice Problems


1. A rope is attached to a bucket filled with water and must pull the bucket vertically up for a distance of 10 feet. You may neglect the weight of the rope and the bucket, but the water initially weighs 100 pounds. The bucket has a hole in it, and at the rate the bucket is being raised, 1.3 pounds of water leaks out for every foot the bucket is lifted. How much work is done in raising the bucket? [935 ft-lb]

2. A rope is attached to a 100 pound weight and must lift the weight vertically a distance of 8 feet. The rope winds around a drum as the weight is raised. The rope itself weighs 0.6 lb/ft and is 50 feet long at the start of the lifting operation. How much work is done? [1021 ft-lb]

3. In optics, the index of refraction is defined by n = sin(i)/sin(r), where "I" is the angle the light makes with respect to the surface of a material it hits, and "r" is the angle the light makes after it refracts into the material. If "I" is increasing at the rate of 0.10 rad/min, what is the rate of change of "r" when I = 15 degrees, if the material has an n = 1.55? [0.0632 rad/min]

4. One of the expressions for electrical power is P = IV, where I is the current and V is the voltage. The power can then be found as the area under the I-V curve. If I=ln(3v)+5V2 , calculate the increased power if V increases from 1 volt to 5 volts.[199.5W]

5.A rectangle is inscribed in the ellipse x2/400 + y2/225 = 1 with its sides parallel to the axes of the ellipse. Find the dimensions of the rectangle of

                    a) maximum area [28.2 x 21.2]

                    b) maximum perimeter [32 x 18]

6. In a modern hotel, where elevators are directly observable from the lobby area, a person in the lobby observes one of the elevators rising at the rate of 12.0 ft/sec. If the person was 50.0 ft from the base of the elevator when it left the lobby, how fast is the angle of elevation of the line of sight to the elevator increasing 10 seconds later? [0.0247 rad/sec]

7. A sky diver falls under the pull of gravity, and the retarding force due to air resistance is 0.3v, where "v" is the velocity. The sky diver weighs 150 pounds, so that the acceleration is given by 150-0.3v. Determine the expression for velocity as a function of time t if v = 0 when t = 0. [ v = 500(1-e-0.3t )]