A 32.4 kg wagon is towed up a hill inclined at 18.4◦ with respect to the horizontal. The tow rope is parallel to the incline and has a tension of 112 N in it. Assume that the wagon starts from rest at the bottom of the hill, and neglect friction. The acceleration of gravity is 9.8 m/s^2. How fast is the wagon going after moving 43.8 m up the hill? Answer in units of m/s
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|dw:1327726001389:dw|By paying attention to the geometry of our situation we see that the force of gravity acting against the tension of the rope is\[|F_g|\sin\theta\]The total force acting on the object along the direction of the incline will be proportional to its acceleration, which we can find by summing the forces along the side of the hill..\[F=T-|F_g|\sin\theta\]\[ma=T-mg\sin\theta\]Assuming 'up the hill' is in the direction along the theta direction we can calculate final velocity from the kinematic equation\[v_f^2=v_o^2+2ad\]In our case vo=0 because the wagon starts from rest.
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