369111517192527 please show work 75 Exercises See CalcChatco

3.6.9.11.15.17.19.25.27

please show work

7.5 Exercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises. Constant Force In Exercises 1-4, determine the work done by the constant force 15. Pumping Water A rectangular tank with a base 4 feet by 1. A 1200-pound steel beam is lifted 40 feet. 2. An electric hoist lifts a 2500-pound car 6 feet. 3. A force of 112 newtons is required to slide a cement block 5 feet and a height of 4 feet is full of water (see figure). The water weighs 62.4 pounds per cubic foot. How much work is done in pumping water out over the top edge in order to empty (a) half of the tank and (b) all of the tank? 8 meters in a construction project. 4. The locomotive of a freight train pulls its cars with a constant 4 ft force of 9 tons a distance of one-half mile. Hooke\'s Law In Exercises 5-10, useHooke\'s Law to determine the variable force in the spring problem. 4 ft 5. A force of 5 pounds compresses a 15-inch spring a total of 3 inches. How much work is done in compressing the s 16. Think About It Explain why the answer in part (b) of 6. A force of 250 newtons stretches a spring 30 centimeters. How Exercise 15 is not twice the answer in part (a). much work is done in stretching the spring from 20 centimeters to 50 centimeters? 17. Pumping Water A cylindrical water tank 4 meters high with a radius of 2 meters is buried so that the top of the tank is 1 meter below ground level (see figure). How much work is done in pumping a full tank of water up to ground level? (The water weighs 9800 newtons per cubic meter.) 7. A force of 20 pounds stretches a spring 9 inches in an exercise machine. Find the work done in stretching the spring 1 foot from its natural position, 8. An overhead garage door has two springs, one on each side of the door. A force of 15 pounds is required to stretch each spring 1 foot. Because of the pulley system, the springs stretch only one-half the distance the door travels. The door moves a total of 8 feet, and the springs are at their natural length when the door is open. Find the work done by the pair of springs Ground leve 9. Eighteen foot-pounds of work is required to stretch a spring 4 inches from its natural length. Find the work required to stretch the spring an additional 3 inches. 10. Seven and one-half foot-pounds of work is required to Figure for 17 Figure for 18 compress a spring 2 inches from its natural length. Find the work required to compress the spring an additional one-half inch. 18. Pumping Water Suppose the tank in Exercise 17 is located on a tower so that the bottom of the tank is 10 meters above the level of a stream (see figure). How much work is done in filling the tank half full of water through a hole in the bottom, using water from the stream? 11. Propulsion Neglecting air resistance and the weight of the propellant, determine the work done in propelling a five-ton satellite to a height of (a) 100 miles above Earth and (b) 300 miles above Earth. 19. Pumping Water An open tank has the shape of a right circular cone (see figure). The tank is 8 feet across the top and 6 feet high. How much work is done in emptying the tank by pumping the water over the top edge? 12. Propulsion Use the information in Exercise 11 to write the work W of the propulsion system as a function of the height h of the satellite above Earth. Find the limit (if it exists) of W as h approaches infinity 13. Propulsion Neglecting air resistance and the weight of the propellant, determine the work done in propelling a 10-ton satellite to a height of (a) 11,000 miles above Earth and (b) 22,000 miles above Earth. 14. Propulsion A lunar module weighs 12 tons on the surface of Earth. How much work is done in propelling the module from the surface of the moon to a height of 50 miles? Consider the radius of themoon to be 1100 miles and its force of gravity to be one-sixth that of Earth. 4 4

Solution

3) force = 12 newtons

work done to slide the cement block to 8 metres

work done by constant force = w = f*d cos theta

since theta = 0 degrees here

therefore , work done = 12*8 * cos 0 = 96 joules

6) force = 250 newtons

work done in stretching the spring from 20 cm to 50 cm

displacement = 30 cm

therefore work done = 250* .30 = 75 joules