A projectile of mass 0.25 kg is at a height of 30 m above horizontal ground and travelling at aspeed of 15 m s –1 . A short time later, it is at a height of 35 m above the horizontal ground andtravelling at a speed of 5.0 m s –1 .How much work is done against air resistance during this time?A 0 J B 13 J C 25 J D 37 J
Question
A projectile of mass 0.25 kg is at a height of 30 m above horizontal ground and travelling at aspeed of 15 m s –1 . A short time later, it is at a height of 35 m above the horizontal ground andtravelling at a speed of 5.0 m s –1 .How much work is done against air resistance during this time?A 0 J B 13 J C 25 J D 37 J
Solution
To solve this problem, we need to use the principle of conservation of energy. The total energy of the projectile at the beginning and at the end of the time interval should be the same, assuming no external work is done.
- Calculate the initial kinetic energy (KE1) and potential energy (PE1) of the projectile.
KE1 = 0.5 * m * v1^2 = 0.5 * 0.25 kg * (15 m/s)^2 = 28.125 J PE1 = m * g * h1 = 0.25 kg * 9.8 m/s^2 * 30 m = 73.5 J
The total initial energy (E1) is the sum of KE1 and PE1, so E1 = 28.125 J + 73.5 J = 101.625 J.
- Calculate the final kinetic energy (KE2) and potential energy (PE2) of the projectile.
KE2 = 0.5 * m * v2^2 = 0.5 * 0.25 kg * (5 m/s)^2 = 3.125 J PE2 = m * g * h2 = 0.25 kg * 9.8 m/s^2 * 35 m = 85.75 J
The total final energy (E2) is the sum of KE2 and PE2, so E2 = 3.125 J + 85.75 J = 88.875 J.
- The work done against air resistance is the difference between the initial and final total energy, so W = E1 - E2 = 101.625 J - 88.875 J = 12.75 J.
So, the closest answer is B) 13 J.
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