pg 155, #30, \"college physics\" serway 8th: A projectile of mass \"m\" is fired
ID: 1692436 • Letter: P
Question
pg 155, #30, "college physics" serway 8th:A projectile of mass "m" is fired horizontally with an initial speed of "Vo" from a height of "h" above a flat, deseert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, Vo, h, and g: A)work done by the force of gravity on the projectile
B) change in kinetic energy of the projectile since it was fired
C) final kinetic energy of the projectile
D)are any of the answers changed if the initial angle is changed?
Explanation / Answer
(a) The formula for work done by gravity is Wgrav = - (change in energy) So, change in energy is = final - initial = 0 - mgh = - mgh So work done by gravity is : Wgrav = - (-mgh) Wgrav = (mgh) J This is true because based on the definition of work done, work done is positive only when the force acting on the object and the object's displacement are in the same direction. Since the projectile is falling and gravity is acting downwards, Wgrav is positive. (b) Using principle of conversation of energy, K1 + U1 = K2 + U2 Rearranging: U1 - U2 = K2 - K1 mgh - 0 = K2 - K1 (RHS is final K.E. - initial K.E.) mgh = change in kinetic energy (c) K1 is given to be 0.5m(Vo^2) Thus, mgh = K2 - K1 K2 = mgh + 0.5m(Vo^2) (d) For this last part, identify the forces acting on the projectile first. Since the only force acting on the projectile is gravity, and air resistance is neglected, we can conclude that the projectile only has conservative forces acting on it. Non-conservative forces are friction , air resistance etc where energy is lost and cannot be gotten back even though I throw the projectile back up to height h. Based on the definition of law of conservation of energy, an object with only conservative forces acting on it do not need to take into account its' path of motion; only consider the final and initial positions of the journey and the path can be ignored. Thus, the energy values, both kinetic and potential, will not change even though the angle is changed.Related Questions
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