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I was able to solve this problem but want to solidify the concept behind it;

A gardener pushes down along a handle of a lawnmower of 20 kg mass with a force of 150 N. The handle makes an angle of 60degrees with the ground. Calculate the instantaneous acceleration of the mower if the frictional force between the mower and the ground at that instant is 25 N.


It makes sense to find the horizontal component of the net force since it is where the friction is occurring. However, how might this be the instantaneous acceleration, wouldn't that require for me to find the time so that I can use a=v/t? or is the fact that the frictional force slowed down the lawnmower immediately sufficient to consider the acceleration immediate.

Please notify me if I require more elaboration in my question. Thank you.

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    $\begingroup$ It might be "bad terminology". Being asked to calculate the instantaneous acceleration $$a=\frac{dv}{dt}$$ gives the impression that the acceleration is variable. But it might well be constant as well. The point is, at a certain time, you have all the values above for the forces, and other parameters, and you are asked to calculate the acceleration with these physical values and conditions. Hence the term "instantaneous". $\endgroup$ – joseph h May 15 at 2:49

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