A 50kg skier pushes off the top of a hill with an initial speed of 5m/s. Neglect friction. How fast will he be moving after dropping 20m in elevation?
I was wondering if i did this problem correct.
PE + KE_f = KE_i
mgh +1/2mv_f^2 =1/2mv_i^2
50(10)(20) + 1/2 (50)v_f^2 = 1/2(50)(5)^2
10000+ 25v_f^2 =625
25v_f^2=-9375
vf^2=-375
vf=19.36 m/s
To solve this problem, you correctly set up the conservation of mechanical energy equation:
PE + KE_final = KE_initial
The potential energy (PE) at the top of the hill is given by mgh, where m is the mass of the skier (50 kg), g is the acceleration due to gravity (10 m/s^2), and h is the height of the hill (20 m). So, the potential energy is 50 * 10 * 20 = 10,000 J.
The initial kinetic energy (KE_initial) is given by 1/2 * m * v_initial^2, where m is the mass of the skier (50 kg) and v_initial is the initial velocity (5 m/s). So, the initial kinetic energy is 1/2 * 50 * 5^2 = 625 J.
Substituting these values into the equation, we have:
10,000 J + 1/2 * 50 * v_final^2 = 625 J
Simplifying the equation, we get:
25 * v_final^2 = -9375 J
At this point, it appears that you made a mistake. You wrote:
25 * v_final^2 = -9375
However, the left side of the equation should be positive since kinetic energy is never negative. So, it should be:
25 * v_final^2 = 9375 J
Dividing both sides of the equation by 25, we get:
v_final^2 = 375 J
To solve for v_final, we take the square root of both sides:
v_final = √(375 J) ≈ 19.36 m/s
Therefore, the skier will be moving approximately 19.36 m/s after dropping 20 m in elevation.