Question

A block of weight w = 25.0 N sits on a frictionless inclined plane, which makes an angle θ = 28.0 ∘ with respect to the horizontal. A force of magnitude F = 11.7 N , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed. 1. The block moves up an incline with constant speed. What is the total work Wtotal done on the block by all forces as the block moves a distance L = 2.70 m up the incline? Include only the work done after the block has started moving at constant speed, not the work needed to start the block moving from rest. 2. What is Wg, the work done on the block by the force of gravity w⃗  as the block moves a distance L = 2.70 m up the incline?

3. What is WF, the work done on the block by the applied force F⃗  as the block moves a distance L= 2.70 m up the incline? 4.What is WN, the work done on the block by the normal force as the block moves a distance L = 2.70 m up the inclined plane?

Answer 1

step:1

Given that

force F=11.7 N

angle =28

distance L=2.7 m

weight W=mg=25 N

step:2

now we find the total work done

the total work done =[Fcos28-Wsin28]*L

                             =[11.7*cos28-25*sin28]*2.7

                             =[10.3-11.74]*2.7

                            =-3.89 J

step:3

the work done by gravity Wg=-25*sin28*2.7

                                        =-31.7 N

step:4

the work done bt applied force WF=11.7*cos28*2.7

                                                =27.9 J

step:5

the work done by normal force WN=mgcos28*L

                                                  =25*cos28*2.7

                                                  =59.6 J