Question #102171
A 9.5 kg block slides down a 13.2 m long ramp inclined at an angle of 30o above the horizontal. At the bottom of the plane, the block transitions to a flat, smooth table that is 2.1 m in length. There is a spring attached to the wall at the end of the flat portion of the track that has spring constant of 256.0 N/m. What is the velocity of the box when it reaches the bottom of the ramp?
1
Expert's answer
2020-02-07T10:30:21-0500

mgh=mv22mgh=\frac{mv^2}{2}


mglsin30°=mv22mgl\sin 30°=\frac{mv^2}{2}



v2=2glsin30°v^2=2gl\sin30°


v=2glsin30°=29.8113.2sin30°=11.38m/sv=\sqrt{2gl\sin30°}=\sqrt{2\cdot 9.81\cdot 13.2\cdot \sin30°}=11.38m/s








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