- #1
AndreaTechGirl
- 6
- 0
Work done by a friction force, block moving up a ramp--question
100kg block moves up on a rough surface, at an incline of 30 degrees, a constant force P=800 N is applied horizontally moving the block a distance of 3 M up the ramp in a time interval of 2 seconds, v1=0.8 m/s, v2=2.2 m/s, what is the work done by the friction force
d-displacement, W-work, F-force, m-mass, g-gravity, v1-initial velocity, v2-terminal velocity
eq1) W=F*cos(30)*d
eq2) F(friction)=(coefficient of kinetic friction)*mg
eq3) W(friction)=F(friction)*d*cos(theta)
eq4) 1/2m(v2)^2-1/2m(v1)^2=F(friction)*d*cos(theta)
Change in KE of the 8block is equal to the work done on block by friction force so, plugging in for eq4 i get 210, it should be negative because its in the opposite direction-right? Is my set up correct? Thanks for your time!
Homework Statement
100kg block moves up on a rough surface, at an incline of 30 degrees, a constant force P=800 N is applied horizontally moving the block a distance of 3 M up the ramp in a time interval of 2 seconds, v1=0.8 m/s, v2=2.2 m/s, what is the work done by the friction force
Homework Equations
d-displacement, W-work, F-force, m-mass, g-gravity, v1-initial velocity, v2-terminal velocity
eq1) W=F*cos(30)*d
eq2) F(friction)=(coefficient of kinetic friction)*mg
eq3) W(friction)=F(friction)*d*cos(theta)
eq4) 1/2m(v2)^2-1/2m(v1)^2=F(friction)*d*cos(theta)
The Attempt at a Solution
Change in KE of the 8block is equal to the work done on block by friction force so, plugging in for eq4 i get 210, it should be negative because its in the opposite direction-right? Is my set up correct? Thanks for your time!