# Work and Friction

1. Oct 17, 2005

### leezak

A 1420-kg car is being driven up a 7.34° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 545 N. A force F is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight W and the normal force FN directed perpendicular to the road surface. The length of the road up the hill is 240 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 203 kJ?

i found the height and then proceeded to find the work of F (keeping the variable in the equation) and then i found the work of the frictional force by multiplying the given force and the height. Then i broke down the weight into two components, x and y. since the work of the normal force and the y component of gravity are both 0 (considering the fact that cos90 = 0) i disregarded those. i then found the work of the y component of gravity by m*g*h... i added all 3 works to equal the net work given and then i solved for F but everytime i get the wrong answer

2. Oct 17, 2005

### hotvette

Maybe you could show us your calculations for the work done by each of the forces?

3. Oct 17, 2005

### leezak

For the work of F i multiplied F*h = F*30.66, for the work of the frictional force 545*30.66 = 16709.7, for the work of gravity m*g*sin7.34*30.66 = 54509.41, and then i added all those and set them equal to 203000 J (because its the net work)... and then i solved for F to get 4298.14... but thats the wrong answer... can you find what i did wrong? thanks!