a) The net work done by gravity on the cyclist is equal to the change in the kinetic energy of the cyclist-bicycle system. Since the cyclist starts from rest, the initial kinetic energy is zero. The final kinetic energy can be calculated using the equation:
KE = (1/2)mv^2
where m is the mass of the cyclist-bicycle system and v is the final velocity of the cyclist.
The work-energy principle states that the net work done on an object is equal to its change in kinetic energy:
W_net = KE_final - KE_initial
Since the initial kinetic energy is zero, the net work done by gravity on the cyclist is equal to the final kinetic energy:
W_net = (1/2)mv^2
To calculate the final velocity, we can use the conservation of energy principle, which states that the total mechanical energy of a system is conserved:
E_i = E_f
where E_i is the initial mechanical energy and E_f is the final mechanical energy. Since the cyclist starts from rest, the initial mechanical energy is equal to the gravitational potential energy:
E_i = mgh
where h is the vertical distance the cyclist travels down the hill. We can use trigonometry to calculate the vertical distance:
h = d*sin(theta)
where d is the distance the cyclist travels and theta is the angle of the hill in radians.
Plugging in the given values:
h = 260sin(4.5pi/180) = 20.05 m
Therefore, the initial mechanical energy is:
E_i = (80 kg)(9.81 m/s^2)(20.05 m) = 15742 J
Since there is no non-conservative work done on the system, the final mechanical energy is equal to the initial mechanical energy:
E_f = E_i = 15742 J
The final mechanical energy can also be expressed as the sum of the final kinetic and potential energies:
E_f = (1/2)mv^2 + mgh
Plugging in the known values and solving for v:
(1/2)(80 kg)v^2 = 15742 J - (80 kg)(9.81 m/s^2)(20.05 m)
v = sqrt[(2(15742 J - 15686 J))/80 kg] = 7.0 m/s
Therefore, the net work done by gravity on the cyclist is:
W_net = (1/2)(80 kg)(7.0 m/s)^2 = 1960 J (to two significant figures)
b) The final velocity of the cyclist is 7.0 m/s, as calculated in part a).