A cannon is fired with an initial horizontal velocity of 20 m/s and initial vertical velocity of 25 m/s. After 3s in the air, th
1 answer:
You might be interested in
The change in internal energy of a system is given by (second law of thermodynamics)
where Q is the heat absorbed by the system and W is the work done on the system.
In order to correctly evaluate the internal energy change, we must be careful with the signs of Q and W:
Q positive -> Q absorbed by the system
Q negative -> Q released by the system
W positive -> W done on the system by the surroundings
W negative -> W done by the system on the surroundings
In our problem, the heat released by the system is
(with negative sign since it is released by the system), and the work done is 
still with negative sign because it is performed by the system on the surrounding, so the change in internal energy is
where Q is the heat absorbed by the system and W is the work done on the system.
In order to correctly evaluate the internal energy change, we must be careful with the signs of Q and W:
Q positive -> Q absorbed by the system
Q negative -> Q released by the system
W positive -> W done on the system by the surroundings
W negative -> W done by the system on the surroundings
In our problem, the heat released by the system is