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Simple Physics Questions

[From: ] [author: ] [Date: 11-07-04] [Hit: ]
What is the potential energy of the object with respect to the ground?12. A 3.25 kg box is at the top of a frictionaless incline as shown in the diagram (note: base of the incline plane is 8.00 m from the base of the building). What is the potential energy of the box with respect to the bottom of the incline?......
8. What power is needed to lift a loaded elevator of 1500 kg ten stories (30 m) in 2.0 min?

9. A sled is pulled with a rope which makes a 40.00 angle with the ground. The average tension in the rope is 100 N.

a) Find the vertical component of the force.

b) Find the horizontal component of the force.

c) The vertical component does not do any useful work. Using the horizontal component, calculate the average power in pulling the sled a distance of 1.00 km in 20.0 min.

10. The maximum output power of an electric motor is 400 W. It is used to operate a conveyer belt carrying sand. The belt is 12 m long and the sand is lifted 4.0 m. What is the maximum mass of sand that can be allowed to flow onto the conveyer belt every minute without overloading the motor? Ignore any friction of the conveyer belt.

11. A 30.0 N object is held 2.40 m above the ground. What is the potential energy of the object with respect to the ground?


12. A 3.25 kg box is at the top of a frictionaless incline as shown in the diagram (note: base of the incline plane is 8.00 m from the base of the building). What is the potential energy of the box with respect to the bottom of the incline?

13. A 55.0-kg woman pulls herself hand-over-hand up a rope. She goes 10.2 m upwards in 25.0 s.

a. How much work does she do?

b. Find her power output.

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Problem 8

Required Power = (1500*9.8*30)/(2*60) = 3675 W = 4.93 HP

Problem 9

a) the vertical component of the force = 100*sin 40 = 64.3 N

b) the horizontal component of the force = 100*cos 40 = 76.6 N

c) Using the horizontal component, calculate the average power in pulling the sled a distance of 1.00 km in 20.0 min = (76.6*1000)/(20.0*60) = 63.8 W

Problem 10

m*9.8*4/60 = 400 or m = 612 kg

Problem 11

PE = 30*2.40 = 72 J

Problem 12

Data not sufficient

Problem 13

a. How much work does she do = 55.0*9.8*10.2 = 5498 J

b. Find her power output = 5498/25.0 = 220 W
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