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18.07.2019 •
Physics
I've got an energy and work problem. the premise of the problem is: a crate with a mass of m is on a ramp that is inclined at an angle of 30° 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. there are a set of subquestions and i've answered them as best as i can, but i don't think i'm right. could someone give me the answers and me understand why i'm wrong? 1. what is the work done on the crate by the applied force f? --i said cos30 * f * displacement 2. 2. what is the work done on the crate by the gravitational force exerted on the crate by earth? --i said m*9.8*tan30 3. 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.) --i said zero 4. what is the work done on the crate by the frictional force fk? --i don't understand this one 5. what is the total force acting on the crate? --i said f, as in the pre-defined force 'f' 6. what is the work done on the crate by the total force? --i don't understand how this is different from the first one
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Ответ:
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
Ответ: