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The diagram below shows a moving, 5.00-kilogram cart at the foot of a hill 10.0 meters high. For the cart to reach the top of the hill, what is the minimum kinetic energy of the cart in the position shown? [Neglect energy loss due to friction.]

User Danatcofo
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5 votes

Answer:

490.5 Joules

Step-by-step explanation:

For something like this, you can use Ug1-K1=Ug2-K2

with Ug 1 being zero because it's at the bottom of the hill and K2 being 0 because it's at the top of the hill and not moving. So it simplifies to K1=Ug2 and we are solving for K1. Ug is the potential energy of gravity which is equal to mass * gravity*height. the mass is 5kg, gravity is 9.81m/s^2 and the height is 10m, So Ug is 490.5. K1 has to equal Ug to overcome the effect of gravity and make it to the top of the hill, so K1 is also 490.5.

If you were accounting for friction the distance traveled in the x direction would matter, but because there is no friction, the only force stopping you from going up the hill is gravity, so it's exactly as hard as going straight up

User Gil Vegliach
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