Answer:
a) 1.53*10^6 J
b) 509399 J
c) 1.02 *10^6 J
d) 62.7 m/s
Explanation:
Given:-
- The mass of rock m = 520 kg
- The Length of the hill , L = 500 m
- The height of the hill, h = 300 m
- The coefficient of friction, u = 0.25
Find:-
(a) If the gravitational potential energy U of the rock–Earth system is zero at the bottom of the hill, what is the value of U just before the slide?
(b) How much energy is transferred to thermal energy during the slide?
(c) What is the kinetic energy of the rock as it reaches the bottom of the hill?
(d) What is its speed then?
Solution:-
- The change in potential energy of the rock due to gravity can be determined by the following relation:
Δ P.E = Uf - Ui
Δ P.E = 0 - m*g*h
Δ P.E = - (520)*(9.81)*(300)
Δ P.E = - 1.53*10^6 J
- There was a loss of potential energy in the rock as it fell down the hill. The potential energy at the top was = 1.53*10^6 J .
- The amount of energy transferred to thermal energy is the amount of work done against friction as the rock slides down the hill with a decent angle (θ). This decent angle can be determined using trigonometric ratios:
sin ( θ ) = h / L
θ = sin^-1 ( 300/500 )
θ = 36.9°
- The frictional force (Ff) can be determined by the contact force (N) between the rock and the hill surface as follows. Using equilibrium equation:
N - m*g*cos(θ) = 0
N = m*g*cos(θ)
Where,
Ff = u*N
Ff = u*m*g*cos(θ)
- The work done by friction (Wf) that is converted to thermal energy is:
Wf = Ff*L
Wf = u*L*m*g*cos(θ)
Wf = 0.25*500*520*9.81*cos(36.9)
Wf = 509399 J
- The kinetic energy at the bottom of the slide can be determined by the Work-Energy principle applied to the rockslide. The initial kinetic energy at top was zero. The change in total mechanical energy is equal to work done against friction.
Δ K.E + Δ P.E = - Wf
K.E - 1.53*10^6 = - 509399
K.E = 1.02 *10^6 J
- The final speed (vf) of the rock at the bottom of the hill can be determined from the kinetic energy relationship as follows:
K.E = 0.5*m*vf^2
1.02 *10^6 = 0.5*52*vf^2
vf = 62.7 m/s
