Answer to Question #95430 in Physics for louie

Question

A 54 kg person starts from rest, skis down a slope of height 50.00m, rises again to a slope of 25.00m, before finally projecting off with a speed of 10.00 m/s at 25 degrees with respect to the horizontal line. Note that the slopes are very slippery and are considered frictionless. She then lands 2.73 s later after projecting off point C. What are the mechanical energy/ies (Kinetic and Potential Energies) present at points A, B and C? What is the velocity of the person in point B? What is the work done by friction from point A to C? How much further from the person’s range is her landing point?

Answer 1

Since there is no friction on the slopes, the energy $E = T + U$ is conserved. Let the heights of the slopes be $h_A = 50 m$ and $h_C = 25m$. Also, $v_C = 10\frac{m}{2}$, $\theta = 25^\circ$, $t = 2.73 s$.

Then, the energies at different points are:

• A: $T_A = 0 J$, $U_A = m g h_A = 26487 J$ (no kinetic energy, since the person is at rest)
• B: $U_B = 0 J$, since the person is at the ground level. By the law of conservation of energy, $E_A = m g h_A = E_B = T_B + U_B$, so $T_B = m g h_A = 26487 J$. $T_B = \frac{m v_B^2}{2} \Rightarrow v_B = \sqrt{2 g h_A} = 31.32 \frac{m}{s}$ .
• C: $T_C = \frac{m v_C^2}{2} = 2700J$ ,$U_C = m g h_C = 13243.5 J$.

The persons landing point is $L = v_c cos\theta t = 24.74 m$ further.