Am I doing these right?
A mass of 3kg rests on a frictionless floor, Ignore friction and wind resistance.
a) if the mass moves 6m to the left,how much work was done due to gravity?
I set my equation as this
w=F*d*cos@
w=(3kg)(9.8m/s^2)(-6m)(cos90)
and I got 0J
b) if the mass then moves 8m up, how much gravitational work was done to it?
I set my equation up as
w=(3kg)(9.8m/s^2)(8m)(cos180)
w=235.2J
c) instead of doing each leg separately, the mass is moved diagonally to the same point (-6m,8m). how much work was done on it by gravity? compare this to your answers in parts a and b. what does that mean regarding the paths taken by the mass?
I set up my equation
w=(3kg)(9.8m/s^2)(10m)(cos135)
w=208J
Am I doing these correctly? If not can you correct my mistakes?
A mass of 3kg rests on a frictionless floor, Ignore friction and wind resistance.
a) if the mass moves 6m to the left,how much work was done due to gravity?
I set my equation as this
w=F*d*cos@
w=(3kg)(9.8m/s^2)(-6m)(cos90)
and I got 0J
b) if the mass then moves 8m up, how much gravitational work was done to it?
I set my equation up as
w=(3kg)(9.8m/s^2)(8m)(cos180)
w=235.2J
c) instead of doing each leg separately, the mass is moved diagonally to the same point (-6m,8m). how much work was done on it by gravity? compare this to your answers in parts a and b. what does that mean regarding the paths taken by the mass?
I set up my equation
w=(3kg)(9.8m/s^2)(10m)(cos135)
w=208J
Am I doing these correctly? If not can you correct my mistakes?
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The only mistake, I notice is that of sign in b) & c).
"Work done by gravity" is ,first of all, NOT the way to word this statement.
"gravity" without further description is for the uneducated (physics-wise :>)
IMHO, always use the appropriate physics QUANTITY name,
U R talking about.
In this case,
"Work done by the FORCE of gravity"
Force is a vector and as such has direction.
The force of gravity always acts down and in this case the mass move UP. The direction of the force is opposite to the direction of the mass's displacement. ie the angle between the force and displacement is not 0° but is 180° which makes the work done by the force of gravity, negative.
"Work done by gravity" is ,first of all, NOT the way to word this statement.
"gravity" without further description is for the uneducated (physics-wise :>)
IMHO, always use the appropriate physics QUANTITY name,
U R talking about.
In this case,
"Work done by the FORCE of gravity"
Force is a vector and as such has direction.
The force of gravity always acts down and in this case the mass move UP. The direction of the force is opposite to the direction of the mass's displacement. ie the angle between the force and displacement is not 0° but is 180° which makes the work done by the force of gravity, negative.