Work A crate with a mass of m is on a ramp that is inclined at an angle of 30deg
ID: 1380069 • Letter: W
Question
Work A crate with a mass of m is on a ramp that is inclined at an angle of 30degree from the horizontal. A force with a magnitude of F directed parallel to the ramp is used to pull the crate with a constant speed up the ramp a distance of d. What is the work done on the crate by the applied force F? What is the work done on the crate by the gravitational force exerted on the crate by Earth? What is the work done on the crate by the normal force, with a magnitude of Fn, exerted on the crate by the ramp? (Hint: recall that the normal force is perpendicular to the surface of the ramp.) What is the work done on the crate by the frictional force Fk? What is the total force acting on the crate? What is the work done on the crate by the total force?Explanation / Answer
1. Work done on the crate by the applied force.
Work done is given by Force times displacement. From the given question, the force has a magnitude F and displacement d, so the force (vector) F3 = F *d.
2. Work done on the crate by the gravitational force exerted on the crate by Earth is given by mgh where m is the mass of the crate, g is the acceleration due to gravity (9.8 m/s2 ) h is the height of the crate from the ground.
3. Work done on the crate by the normal force Fn
The gravitational force Fg is actually mg. The force mg is resolved into horizontal component x which is mg sin 30. and y component is mg cos 30. The normal force which is of same magnitude but opposite of the y component is given as mg cos 30 and so the work done is given as force times displacement which is mg cos 30 times height of the crate from the base/ground.
4. Work done on the crate by the frictional force Fk
Work done by the frictional force is the product of the frictional force and the displacement of the crate in the direction of the force.
The magnitude of the force in the direction is the x component of the gravitational force which is mg sin 30 and therefore the work done is force times displacement = mg sin 30 times d.
5. Total force acting on the crate: To find the force acting on the crate find the x and y components of each vector and take the resultant.
x component of F is F, y component is 0.
x component of Fn is Fn cos 90 which is 0 and y component is Fn sin 90 which is Fn
x component of Fk is -Fk cos 90 which is 0 and y component is -Fk sin 90 which is -Fk
x component of Fg is -Fg cos 60 which is -0.5 Fg and y component is -Fg sin 60 which is -0.866 Fg
The total force acting on the crate is square root((0.5Fg)2 + (Fn -Fk -0.866 Fg )2 ).
6. Work done by the total force on the crate is total force times displacement.
square root((0.5Fg)2 + (Fn -Fk -0.866 Fg )2 ) times d.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.