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

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To push a 7.5kg shopping cart up a 13-degree incline with an acceleration of 1.41 m/s², the total force required 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, while the gravitational force component is 16.53N. Adding these forces together results in a total force of 27.03N needed to move the cart up the incline. A diagram was referenced to clarify the calculations. This approach effectively breaks down the forces involved in pushing the cart uphill.
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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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