A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25°above the horizontal.Just before it hits the ground its speed is 150 m/s.Over the entire trip the change in the thermal energy of the projectile and air is:
A)+6300 J
B)-6300 J
C)+44,000 J
D)-44,000 J
E)0 J

Question

Quizplus · Accepted Answer

The change in thermal energy can be calculated using the principle of conservation of energy. Initially, the projectile has kinetic energy given by $ \frac{1}{2}mv^2 $, where $ m = 5 $ kg and $ v = 200 $ m/s. Just before it hits the ground, its kinetic energy is given by $ \frac{1}{2}mv^2 $ with $ v = 150 $ m/s. The change in kinetic energy is the difference between these two values, which is converted into thermal energy.Initial kinetic energy: $ \frac{1}{2} \times 5 \times (200)^2 = 100,000 $ JFinal kinetic energy: $ \frac{1}{2} \times 5 \times (150)^2 = 56,250 $ JChange in kinetic energy (which equals the change in thermal energy since energy is conserved): $ 100,000 - 56,250 = 43,750 $ J, which is approximately $ +44,000 $ J.

A 5-Kg Projectile Is Fired Over Level Ground with a Velocity