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The question is this:

Darla pushes a $5.0kg$ crate up a $2.0m$ high $20^o$ frictionless slope by pushing it with a constant horizontal force of $25N$. What is the speed of the crate as it reaches the top of the slope? The first instruction is to solve the problem using work and energy, and the second part says to solve using Newton's laws. So here is my attempt

a) Solve this problem using work and energy

I found that the length traveled up the slope is

$$sin20^o={2.0\over L}$$ $$L={2.0\over sin20^o}=5.9m$$

Then I used the equation $W=F\cdot s$ $$W=25N\cdot(5.9)$$ $$W=148J$$

Then $$W_{tot}=K_2-K_1$$ $$W_{tot}={1\over 2} mv_2^2-{1\over 2} mv_1^2$$ $$148J={1\over 2}5kg(v_2^2)-{1\over 2}5kg(0^2)$$ $$148J=2.5kg(v_2^2)$$ $$v=7.7m/s$$

So I did that part, but when I used Newton's laws I got a different answer:

$$mg=(5.0kg)(9.8m/s^2)=49N$$ y-component of the weight $49(cos20)=46N$ I also put the normal force, $n=46N$ x-component of the weight $49(sin20)=17N$

I chose a coordinate system of the y-component of weight being negative, and the x-component of the weight negative, so I calculated the net force

$$\Sigma F_x=25N-17N=8N$$ $$\Sigma F_y=49N-49N=0$$ $$\Sigma F_{net}=8N$$

Then I attempted to solve for acceleration:

$$F=ma$$ $$8N=(5.0kg)a$$ $$a=1.6m/s^2$$

I used that with a kinematics equation:

$$v^2-v_o^2=2a(x-x_o)$$ $$v^2=2(1.6)(5.9)=19$$ $$v=4.4m/s$$

So obviously this isn't the same answer as the one I got before. Where did I go wrong?

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closed as off-topic by John Rennie, ACuriousMind, user36790, Kyle Kanos, Gert Nov 6 '15 at 17:15

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  • $\begingroup$ in the total work you forgot to add the cosine of the angle between the force and the displacement. Also forgot to add potential energy. $\endgroup$ – user83548 Nov 6 '15 at 0:31
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Problem solved using work and energies

a) Problem solved in Mathcad using work and energy

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  • $\begingroup$ Again unexplained minuses. If the demonstration is wrong please correct it. $\endgroup$ – Energizer777 Nov 9 '15 at 9:59

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