How fast is the wagon going after moving 44.9 m up the hill in the following case:
A 42.4 kg wagon is towed up a hill inclined at 15.6 degrees with respect to the horizontal. The tow rope is parallel to the incline and has a tension of 135 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.
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The mass of the wagon that is being towed up is 42.4 kg. The hill is inclined at an angle 15.6 degrees to the horizontal. The force that is used to pull the wagon up the slope is the tension in the rope and is equal to 135 N.
The gravitational force of attraction that acts vertically downwards on the wagon is 42.4*9.8 = 415.52 N. The component of this parallel to the incline is 415.52*sin 15.6 = 111.74 N and acts in a downwards direction.
This gives the net force acting on the wagon as 23.26 N acting upward. The acceleration due to this force is 0.548 m/s^2. The velocity of the wagon after traveling for 44.9 m up the hill is v where v^2 = 2*0.548*44.9 = 49.2104. This gives v = 7.01 m/s
The required velocity of the wagon is 7.01 m/s
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