The 0.2 kg box below slides down a curved ramp, jumps a small gap and lands on a
ID: 1491555 • Letter: T
Question
The 0.2 kg box below slides down a curved ramp, jumps a small gap and lands on a flat platform. At the point on the ramp shown it is 1.5 m above the floor and its speed is 2.0 m/s. At the point shown on the platform the box is 0.4 m above the floor and sliding at 4.2 m/s. a) How much energy has been converted to non-mechanical form (mostly thermal/internal energy) as the box moves between the two points shown? b) Further down the ramp the object comes to rest. How much energy has been converted to non-mechanical form as the box moves between the initial point shown and the point where the box stops?
Explanation / Answer
If you have a graph in your problem would be clearer, in the diagram is an outline of what I understood from your writing.
Part a)
We seek energy between the start and end points
Emi = K + U = ½ m v2+ mg h
Emi = ½0.2 22 + 0.2 9.8 1.5
Emi = 3.34 J
Emf = K + U = ½ m vf2 + m g hf
Emf = ½ 0.2 4.22 + 0.2 9.8 0.4
Emf = 2.548 J
Em = Emf – Emi
Em = 2.548 – 3.34
Em = -0.792 J
In summary 0.792 J have become to heat
Part b)
An object down the ramp from rest and when it reaches the platform is at rest
We calculate the energy at each point
Emi = K + U
Emi = 0 + mg h2
Emi = 0.2 9.8 1.5
Emi = 2.94 J
Emf = K + U
Emf = 0 + m g h2
Emf = 0.2 9.8 0.4
Emf = 0.784 J
Em = Emf – Emi
Em = 0.784 – 2.94
Em = -2.156 J
In resume 2,156 J have been converted to not mechanical energy
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