A 50.0 kg skier starts at A with an unknown initial speed, skis over a hill which is part of a circle?
A 50.0 kg skier starts at A with an unknown initial speed, skis over a hill which is part of a circle of radius r = 14.8 m (the top of which is at B, h = 4.00 m vertically above A), before skiing down to point C (which is vertically at the same height bottom of the circle) as shown in the figure. Friction is negligible between points A and C. The skier then encounters a rough patch l = 53.0 m long between points C and D with coefficient of kinetic friction \mu_k = 0.275. Since the rough patch fails to bring the skier to rest, shortly after point D the skier collides with a huge uncompressed spring with spring constant k = 1820 N/m. (Figure not drawn to scale!)
Wgravity= -1.96 kJ
1.If the normal force the skier experiences at B (at the crest of the hill) is 5/6 of her weight, what was the speed v_A of the skier at A (i.e. initially).
2.What is the speed v_C at point C, at the bottom of the hill?
3.What is maximum compression in the spring? (i.e. when the skier is finally, but only instantaneously brought to rest after passing through the rough patch and colliding with the spring.)
diagram on this website
for some reason the answers you gave me are not working i cant figure out the problem can u help
i think the weight was suppose to be 5/6
- David GeelanLv 59 years agoFavorite Answer
The normal force at B will be the weight force minus the centripetal force. The weight force is mg, so the centripetal force must be mg/6 = (50*9.8)/6 = 81.7 N
For circular motion the centripetal force is given by F = mv^2/r, so 81.7 = 50v^2/14.8
Rearranging gives v^2 = 24.2 therefore v = 4.9 m/s
Now this is the velocity at B, but we're asked for the velocity at A. Since the skier has risen by 4 m, some of the initial kinetic energy at A will have been converted to gravitational potential energy.
PE = mgh = 50*9*4 = 1800 J
This will be equal to the *additional* kinetic energy the skier had at A, over and above that represented by the 4.9 m/s at B
KE = 1800 = 1/2*mv^2 = 25 v^2 ----> v^2 = 72 ---> v = 8.5 m/s
Add this to the 4.9, and the velocity of the skier at A is 13.4 m/s
At C, the opposite process has happened - some gravitational potential energy has been converted to kinetic energy. Let's consider the situation relative to B (where the velocity is 4.9 m/s). The circle has radius 14.8 m, so it has diameter 29.6 m. The gravitational potential energy difference between B and C is therefore PE = mgh = 50*9.8*29.6 = 14,504 J.
This will be equivalent to the gain in kinetic energy: KE = 1/2 mv^2: 14,504 = 1/2*50*v^2 ---> v = 24.1 m/s. Add this to the 4.9 m/s at B for a total velocity at C of 29 m/s.
The frictional force will cause an acceleration in the opposite direction to the motion (a deceleration).
The frictional force is given by F(fric) = mu(k) * F(norm), but in this case the normal force is just the weight force of the skier. F(fric) = 0.275*50*9.8 = 134.8 N.
This will cause acceleration in accordance with Newton's second law, a = F/m = 134.8/50 = 2.7 m/s/s
The final velocity will be given by v^2 = u^2 + 2ad (where a will be negative. Some places use 's' instead of 'd' for distance)
v^2 = 29^2 - 2*2.7*53 = 841 - 286.2 = 554.8 ----> v = 23.6 m/s
When the skier hits the spring, the kinetic energy of this motion will be converted to spring potential energy in the spring:
Kinetic energy: KE = 1/2 mv^2 = 1/2*50*23.6^2 = 13,924 J
Spring potential energy: SPE = 1/2 k x^2 ----> 13,924 = 1/2*1820*x^2
Rearranging gives x^2 = 15.3 ---> x = 3.9 m
All the best with your studies!