what is the total work wf done on the block by the applied force f as the block moves a distance

What is the total work wf done on the block by the applied...

Transcript

00:01 Hello and welcome to this video solution of numerate so here we are given that there is a block of mass 1 kg that is placed on an inclined plane, right? so i take mass m to be equal to 1 kg this force f that is pulling this block up the incline is 70 newton right the coefficient of kinetic friction between the plane and the block is 0 .5.

00:30 So mu i take equal to 0 .5 right and the angle of incline theta is 50 degrees, right? so based on this you have to calculate the total work done on the block by the force of friction as the block moves 10 meters to the incline so here you have to calculate the work done due to friction, right? so first of all, you have to find out the force of friction then the weight mg is acting downwards there is two component of weight one perpendicular to the incline which is mg cos theta right and this is mg sine theta and this is the normal reaction in and if you see since it is moving up the frictional force will act downwards so the frictional force fr will be equal to mu times the normal reaction, which is mu mg cos theta because this normal reaction will balance this component of the weight perpendicular component, right so work done due to friction will be equal to this is mu mg cos theta is a force right fr times the displacement right now displacement is in opposite direction of the friction so here i take minus l, right? so you can plug in the values minus this is 0 .5 mass is 100 kg 100 into g is 9 .8 sorry, mass is 1 kg into 9 .8 cos of this angle theta is 50 degrees times the length is 10 meters, right? so, let's see how much we have here now this is 31 .5 newton 5 joules is a work done due to friction, right? next you have to calculate the work done due to this force f up the incline, right wf it's pretty simple so it's f times length because now this force is acting in same direction as a displacement, right? so it's a 700 joules...

Question

What is the total work WF done on the block by the applied force F⃗ as the block moves a distance 10.0 m up the incline?

Question

What is the total work WF done on the block by the applied force F⃗ as the block moves a distance 10.0 m up the incline?

Please give Ace some feedback