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I've been trying so long at this problem to no avail. I drew my free body diagram, but I'm unsure which formula to use. enter image description here

Could someone help me out?

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2 Answers 2

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So, we can start saying that the theorem tell us that the variation of the KE is equal to the work done.

Remember that W=F*s and KE=(1/2)(m)(v)^2

Now you have to decompose the weight force (you will see that with this decomposing , is formed a right triangle) , in this way you can obtain the value of that force in her parallel component. In fact the weight force is 98,1 N; to obtain the parallel component:

                  Fp=(98.1N)(sin35)=56N 

Now we can calcolate the work done (remembering to trasform from cm in m, so from 20cm to 0,20m)

                 W=(Fp)(s)=(56N)(0,20m)=11 J

We can pass to considerer the KE. In a first time we have a

KE1=(1/2)(m)(v1)^2=320 J

for the second time , or rather the point after percorring 0,20m we have the unknown velocity. To obtain this one we use the theorem

11J=320J-5kg(v)^2 so v=sqrt[(11J-320J)/(-5kg)]=7,86 m/s

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Hint: Think about dividing the gravitational force into components, with one in the same line as the velocity, and the other perpendicular to it.

Apply the Work Energy Theorem after that. (i.e Work Done (The component of the force, multiplied with the given distance (i.e. $20$ cm) by the gravitational force along the displacement is equal to the change in K.E.)

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