How much force is needed to push a shopping cart up an incline?

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SUMMARY

The force required to push a 7.5kg shopping cart up a 13-degree incline with an acceleration of 1.41 m/s² is calculated by combining the force needed for acceleration and the gravitational component acting down the slope. The force for acceleration on level ground is 10.5N (7.5kg * 1.4m/s²). The gravitational force component is 16.53N (7.5kg * 9.8m/s² * sin(13°)). Therefore, the total force needed is 27.03N.

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A shopper pushes a 7.5kg shopping cart up a 13 degree incline.



I need to find the force required to give the cart an acceleration of 1.41m\s\s up the incline.



ive tried breaking the weight, force and acceleration into components but i just end up getting more confused...any help on this would be great, thanks for your time.
 
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F=ma

If it was on level ground and you wanted it to go at 1.4ms-2, the force needed would be just 7.5*1.4=10.5N

Now because it's on an incline you have to add in that component also. This diagram should explain it:

http://instruct.tri-c.edu/fgram/web/Image86.gif

You need to counteract the mgsin0 by supplying that force in the opposite direction - up the slope.

So that is 7.5*9.8*Sin13 = 16.53N

Add the two together and you get 27.03N

Eamon
 
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Wow, now i understand! thanks so much for your help!
 

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