- #1
ac7597
- 126
- 6
- Homework Statement
- Hetsut is the foreman of a construction project in ancient Egypt. He needs to move a giant block of stone, of mass 12 metric tons, from the docks to the temple grounds. He can go along the roads by travelling 295 meters east, then 89 meters north. Along the roads, the coefficient of kinetic friction between block and ground is 0.12.
"Excuse me, honored foreman," says one of the workmen, "but would it be better to take a short cut along a direct route, across the fields?" The foreman thinks for a moment -- he knows that the coefficient of kinetic friction between stone and dirt in the fields is higher, about 0.22.
How much work would the team have to do on the block to move it from rest at the docks to a position at rest at the temple grounds, if they go along the roads?
How much work would the team have to do on the block to move it from rest at the docks to a position at rest at the temple grounds, if they go straight across the fields?
- Relevant Equations
- work of gravity=mgH
12 metric ton=12000kg
straight length=(295^2+89^2)^(1/2)=308.13m
Homework Statement: Hetsut is the foreman of a construction project in ancient Egypt. He needs to move a giant block of stone, of mass 12 metric tons, from the docks to the temple grounds. He can go along the roads by traveling 295 meters east, then 89 meters north. Along the roads, the coefficient of kinetic friction between block and ground is 0.12.
"Excuse me, honored foreman," says one of the workmen, "but would it be better to take a short cut along a direct route, across the fields?" The foreman thinks for a moment -- he knows that the coefficient of kinetic friction between stone and dirt in the fields is higher, about 0.22.
How much work would the team have to do on the block to move it from rest at the docks to a position at rest at the temple grounds, if they go along the roads?
How much work would the team have to do on the block to move it from rest at the docks to a position at rest at the temple grounds, if they go straight across the fields?
Homework Equations: work of gravity=mgH
12 metric ton=12000kg
straight length=(295^2+89^2)^(1/2)=308.13m
work=mgH-u(mgH).
work to east=(12000kg)(9.8m/s^2)(295m)-(0.12)(12000kg)(9.8m/s^2)(295m)=30.5E6 J
work to north=(12000kg)(9.8m/s^2)(89m)-(0.12)(12000kg)(9.8m/s^2)(89m)=9.21E6 J
total work= 30.5E6 J+ 9.21E6 J= 39.7E6 J
work to straight=(12000kg)(9.8m/s^2)(308.13m)-(0.22)(12000kg)(9.8m/s^2)(308.13m)=28.3E6 J
"Excuse me, honored foreman," says one of the workmen, "but would it be better to take a short cut along a direct route, across the fields?" The foreman thinks for a moment -- he knows that the coefficient of kinetic friction between stone and dirt in the fields is higher, about 0.22.
How much work would the team have to do on the block to move it from rest at the docks to a position at rest at the temple grounds, if they go along the roads?
How much work would the team have to do on the block to move it from rest at the docks to a position at rest at the temple grounds, if they go straight across the fields?
Homework Equations: work of gravity=mgH
12 metric ton=12000kg
straight length=(295^2+89^2)^(1/2)=308.13m
work=mgH-u(mgH).
work to east=(12000kg)(9.8m/s^2)(295m)-(0.12)(12000kg)(9.8m/s^2)(295m)=30.5E6 J
work to north=(12000kg)(9.8m/s^2)(89m)-(0.12)(12000kg)(9.8m/s^2)(89m)=9.21E6 J
total work= 30.5E6 J+ 9.21E6 J= 39.7E6 J
work to straight=(12000kg)(9.8m/s^2)(308.13m)-(0.22)(12000kg)(9.8m/s^2)(308.13m)=28.3E6 J
