Here in this case we can use work energy theorem
As per work energy theorem
Work done by all forces = Change in kinetic Energy of the object
Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.
Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

also we know that


Now kinetic energy is given by





Now by work energy theorem
Work done = 10500 - 0 = 10500 J
So in the above case work done on sphere is 10500 J
Answer:
See explanation below
Explanation:
The equation to use for this is the following:
dU = q + w
As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.
Therefore the change in the energy would be:
dU = -2.59x10^4 - 6.46^4
dU = -9.05x10^4 kJ
Answer:
1÷60 h
time equals distance upon speed
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