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calculus: finding total work done?
A box is dragged 8m by a 40 N force applied at an angle of 25° to the level ground. It is then dragged to the top of a 5 m ramp by the same force. If the ramp is inclined at 12◦ to the ground, find the total work done.
1 Answer
- Ron WLv 78 years agoFavorite Answer
This is a physics question. You would probably get a more prompt response if you posted this to the Physics forum.
There's no mention of friction so I assume we ignore it.
First part:
The forces on the box are the dragging force, acting at an angle of 25° with respect to the horizontal; gravity, acting vertically downward; and the normal force, acting vertically upward. Resolve the dragging force into horizontal and vertical components.
There is no acceleration in the vertical direction, so the net vertical force on the box is zero; hence, there is no work done in the vertical direction.
The net force in the horizontal direction is the horizontal component of the dragging force, which has magnitude 40 cos(25°). The work done in this part is the force times the distance through which the force acts, which is (40 cos(25°)N)(8m) = 320 cos(25°) J.
For the second part (up the ramp), resolve forces into components parallel and perpendicular to the ramp. There is work done only by the component of force parallel to the ramp.
Can you take it from there?
