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Box pushed up incline
An 80 N box is pulled 20 m up a 30 degree incline by an applied force of 100 N that points upwards parallel to the incline. If the coefficient of kinetic friction between the box and incline is .220, what is the change in kinetic energy of the box?
Customer support service by UserEcho
Ok we have 2 forms of potential, kinetic and gravitational potential, and two sources of work, the driving force and friction.
The driving force is what is giving the box energy, so let's start with that. It gives The box energy so the work it does is positive. W= F*d. These values are given, so that part is easy.
Next, it takes energy to go up the ramp because gravitational potential is increasing. U=mgh, so subtract that from the work done by the driving force.
The last piece is the work done by friction. This takes energy from the block. First you have to figure out the force of friction. This can be done with Newton's Secind law, and is much easier if you draw a diagram! Again, W =F*d. We know the work should be negative, and it is because F&d point in opposite directions. Subtract this from our running energy and you are left with kinetic energy.
In an equation, this looks like KE+U_grav=W_drive+ W_friction
Hope this helps!