Posted by Leah on .
An object is placed on an inclined plane which forms an angle B = 18º with the horizontal. An external force F of magnitude 14.3 N is applied to the object. The force F acts in a direction which forms an angle A = 62º with the horizontal. As a result of the force F, the object moves a distance of 0.8 m along the inclined plane. What work was done by the source of force F?
I tried solving this using by drawing it out, using trig and newton's laws to calculate Fx and Fy and used these values to calculate work in x-direction and y-direction.
Finally I used Workx and Workx to find overall work with pythagoras....is this correct?
I got 6.21J as the work
No on the workx +worky. Work is not a vector. You just add them as scalars.