Question

Compute the work done by a 100 lb force pushing a box 10 feet up a 25 degree incline. Assume that the force is parallel with the incline.

Answer 1

Anyone? :/

Answer 2

You have to approach this by examining the work in the vertical direction and the horizontal direction. You also have to determine the vertical and horizontal distances by using trigonometry to solve for the vertical and horizontal distances. The incline is 10 feet, so the horizontal distance is given by 10 cos(25)=9 and the vertical distance is given by 10 sin(25)=4.2. You now have to find the force applied in each direction. Because the force is along the hypotenuse, you find the horizontal force and vertical force in a similar manner. The horizontal force is given by 100 cos(25)=90.6 and the vertical is given by 100 sin(25)=42.3. Work is given by W=Fd. So the work in the horizontal direction is 9*90.6=815.4 and the vertical work is 177.66. Now we take these components to find the work in the direction of the incline which is the magnitude of the components. This is given by \[\sqrt{815.4^2+177.66^2}=834.5\] I'm not entirely positive about this, so I apologize if I approached this incorrectly.

Answer 3

I think it's just 100 times 10 since the force is parallel to the incline, and for the box to move at all the force pushing up the incline has to equal the force pushing down the incline.

The force of friction and force of box parallel to the incline and pushing down the incline

Since we have no information on the coefficient of friction between the incline and the box, nor do we have any information on the mass of the box, the 25 degrees seems useless to me.

Answer 4

Thxs for the replies!!!!!!