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1. A force of 10,000 N is applied to a stationary wall. How much work is performed?. W = F X D. 1. A force of 10,000 N is applied to a stationary wall. How much work is performed?. W = F X D. = 10,000 N X 0 m.
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1. A force of 10,000 N is applied to a stationary wall. How much work is performed? W = F X D
1. A force of 10,000 N is applied to a stationary wall. How much work is performed? W = F X D = 10,000 N X 0 m
1. A force of 10,000 N is applied to a stationary wall. How much work is performed? W = F X D = 10,000 N X 0 m = 0 Nm
1. A force of 10,000 N is applied to a stationary wall. How much work is performed? W = F X D = 10,000 N X 0 m = 0 Nm = 0 J
2. A 950-N skydiver jumps from an altitude of 3000 m. What is the total work performed on the skydiver? W = F X D
2. A 950-N skydiver jumps from an altitude of 3000 m. What is the total work performed on the skydiver? W = F X D = 950 N X 3000 m
2. A 950-N skydiver jumps from an altitude of 3000 m. What is the total work performed on the skydiver? W = F X D = 950 N X 3000 m = 2,850,000 J
3. You are walking from your math class to your English class. You are carrying books that weigh 20 N. You walk 45 m down the hall, climb 15 meters up the stairs and then walk another 30 m to your English class. What is the total work done on the books? W = F X D
3. You are walking from your math class to your English class. You are carrying books that weigh 20 N. You walk 45 m down the hall, climb 15 meters up the stairs and then walk another 30 m to your English class. What is the total work done on the books? W = F X D = 20 N X 15 m
3. You are walking from your math class to your English class. You are carrying books that weigh 20 N. You walk 45 m down the hall, climb 15 meters up the stairs and then walk another 30 m to your English class. What is the total work done on the books? W = F X D = 20 N X 15 m = 300 J
4. How much work is done by a crane that lowers 1000 N of material a distance of 150 m? W = F X D
4. How much work is done by a crane that lowers 1000 N of material a distance of 150 m? W = F X D = 1000 N X 150 m
4. How much work is done by a crane that lowers 1000 N of material a distance of 150 m? W = F X D = 1000 N X 150 m = 150,000 J
5. A weightlifter lifts a 150 kg barbell above his head from the floor to a height of 2 m. He holds the barbell there for 5 seconds. How much work does he do during that 5 second interval? weight = M X 9.8 N kg
5. A weightlifter lifts a 150 kg barbell above his head from the floor to a height of 2 m. He holds the barbell there for 5 seconds. How much work does he do during that 5 second interval? weight = M X 9.8 N kg = 150 kg X 9.8 N kg
5. A weightlifter lifts a 150 kg barbell above his head from the floor to a height of 2 m. He holds the barbell there for 5 seconds. How much work does he do during that 5 second interval? weight = M X 9.8 N kg = 150 kg X 9.8 N kg = 1470 N
5. A weightlifter lifts a 150 kg barbell above his head from the floor to a height of 2 m. He holds the barbell there for 5 seconds. How much work does he do during that 5 second interval? W = F X D
5. A weightlifter lifts a 150 kg barbell above his head from the floor to a height of 2 m. He holds the barbell there for 5 seconds. How much work does he do during that 5 second interval? W = F X D = 1470 N X 0 m
5. A weightlifter lifts a 150 kg barbell above his head from the floor to a height of 2 m. He holds the barbell there for 5 seconds. How much work does he do during that 5 second interval? W = F X D = 1470 N X 0 m = 0 J
6. A 5 kg rock is lifted 2 m in 5 sec. How much work is done?
6. A 5 kg rock is lifted 2 m in 5 sec. How much work is done? weight = M X 9.8 N kg = 5 kg X 9.8 N kg = 49 N
6. A 5 kg rock is lifted 2 m in 5 sec. How much work is done? W = F X D
6. A 5 kg rock is lifted 2 m in 5 sec. How much work is done? W = F X D = 49 N X 2 m
6. A 5 kg rock is lifted 2 m in 5 sec. How much work is done? W = F X D = 49 N X 2 m = 98 J
7. A teacher pushed a 10-kg desk across a a floor for a distance of 5 m. She exerted a horizontal force of 20 N. How much work was done? W = F X D
7. A teacher pushed a 10-kg desk across a a floor for a distance of 5 m. She exerted a horizontal force of 20 N. How much work was done? W = F X D = 20 N X 5 m
7. A teacher pushed a 10-kg desk across a a floor for a distance of 5 m. She exerted a horizontal force of 20 N. How much work was done? W = F X D = 20 N X 5 m = 100 J
8. If 4,000 Nm are used to raise a 294 N object . How high is the mass raised? W = F X D
8. If 4,000 Nm are used to raise a 294 N object . How high is the mass raised? W = F X D 4,000 Nm= 294 N X D
8. If 4,000 Nm are used to raise a 294 N object . How high is the mass raised? W = F X D 4,000 Nm= 294 N X D 4,000 Nm= 294 N X D 294 N 294 N
8. If 4,000 Nm are used to raise a 294 N object . How high is the mass raised? W = F X D 4,000 Nm= 294 N X D 4,000 Nm= 294 N X D 294 N 294 N D = 13.6 m
8. Alternate solution D = W F D= 4000 Nm 294 N D = 13.6 m
9. A bulldozer performs 80,000 Nm of work pushing dirt a distance of 16 m. What is the force of the dirt? W = F X D
9. A bulldozer performs 80,000 Nm of work pushing dirt a distance of 16 m. What is the force of the dirt? W = F X D 80,000 Nm = F X 16 m
9. A bulldozer performs 80,000 Nm of work pushing dirt a distance of 16 m. What is the force of the dirt? W = F X D 80,000 Nm = F X 16 m 80,000 Nm = F X 16 m 16 m 16 m
9. A bulldozer performs 80,000 Nm of work pushing dirt a distance of 16 m. What is the force of the dirt? W = F X D 80,000 Nm = F X 16 m 80,000 Nm = F X 16 m 16 m 16 m F = 5,000 N
F = W D 9. Alternate solution F= 80,000 Nm 16 m F = 5000 N
10. An ant does 1 N-m of work dragging a 0.0020-N grain of sugar. How far does the ant drag the sugar? W = F X D
10. An ant does 1 N-m of work dragging a 0.0020-N grain of sugar. How far does the ant drag the sugar? W = F X D 1 Nm= 0.0020 X D
10. An ant does 1 N-m of work dragging a 0.0020-N grain of sugar. How far does the ant drag the sugar? W = F X D 1 Nm= 0.0020 X D 1 Nm= 0.0020 N X D 0.0020 N 0.0020 N
10. An ant does 1 N-m of work dragging a 0.0020-N grain of sugar. How far does the ant drag the sugar? W = F X D 1 Nm= 0.0020 X D 1 Nm= 0.0020 N X D 0.0020 N 0.0020 N D = 500 m
10. Alternate solution D = W F D = 1 Nm 0.0020 N D = 500 m