Question

A 2 kg rock is fired at an angle of 60 degree above the horizontal with an initial speed 20 m/s by someone standing on the edge of a cliff 40 m above level ground. The initial position of the rock is taken to be y = 0. Air resistance is not neglected in this situation!

(a) The initial mechanical energy of the rock is?
(b) The work being done on the rock by the force of gravity after being fired and moving up through the air.
(c)At the highest point in the rock's trajectory, its mechanical energy.
(d) The work being done on the rock by air resistance as it falls down to the ground.

Answer 1

(a)

h = height at the point of launch = 40 m

m = mass of the rock = 2 kg

v_i = speed of launch = 20 m/s

Initial mechanical energy is given as

TE_i = mgh + (0.5) m v_i²

TE_i = (2 x 9.8 x 40) + (0.5) (2) (20)²

TE_i = 1184 J

b)

At the highest point :

v_f = final speed = v_i Co60 = 20 Cos60 = 10 m/s

W = work done by gravity

Using conservation of energy

W = Change in kinetic energy

W = Final Kinetic energy - Initial kinetic energy

W = (0.5) m (v_f² - v_i²) = (0.5) (2) ((10)² - (20)²) = - 300 J

c)

Mechanical energy at the highest point is given as

TE_f = TE_i = 1184 J