In the usual analysis, work needed to lift an object of mass (m) a certain distance (h) is the same whether it is through lifting the object straight up (W=mgh) or pushing it up a ramp with length d and angle A (W = mgsinA x h/sinA = mgh). However, this does not account for the work needed to move the said object the ground distance (h/tan A), no? For example, in a 30/60/90 degree triangle scenario with height of 1 meter ramp 2 meter, and ground distance square root of 3 meter, the work needed to move an object of 2 kg horizontally first would be 2gxsquare root of 3 and the work to move the object up 1 m would be 2g. The combined work would be 2g (square root of 3 + 1), no? The usual analysis of moving that object up the 2m ramp is 2gx(1/2) divided by 1/(1/2) = g x 2 = 2g. These two methods don't add up. What am I missing? Is the usual ramp analysis wrong, in that mgxsin A or 2gx(1/2) or g can't the force needed to move the object up the 2 meter ramp? Should that force be (1+square root of 3)g instead? What am I missing? Thank you so very much for helping out!
asked Mar 7, 2017 at 4:15
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$\begingroup$ What is the work necessary to move an object horizontally a distance $d$? $\endgroup$
Commented Mar 7, 2017 at 4:21
$\begingroup$ I think I got it. Work is NOT necessary to move the object horizontally assuming constant velocity and absence of friction. This also assumes that negligible acceleration and deceleration to get the object to start moving and stopping over the horizontal distance. Therefore, there is no horizontal work component, right? $\endgroup$
Commented Mar 7, 2017 at 5:28