What is the skier's initial velocity?

AI Thread Summary
The skier's initial velocity problem involves calculating the skier's speed at the top of a hill using principles of physics. The skier, weighing 50.0 kg, experiences a normal force of 5/6 of their weight at the hill's peak, which is 4.00 m high. The centripetal force is derived from the gravitational force, leading to an initial calculation of the skier's speed at the top as 10.92 m/s. However, the conservation of energy approach indicates that the initial velocity should be recalculated, as the first part's velocity is incorrect. The discussion emphasizes the importance of considering all forces acting on the skier and suggests that the top-of-the-hill velocity should be less than half of the initially calculated value.
Virtuozzo
Messages
1
Reaction score
0

Homework Statement



A 50.0kg skier starts at A with an unknown initial velocity. The skier then goes over a hill (point B) which is 4.00m above point A. At point B (top of the hill), the skier experiences a Force normal 5/6 of his/her weight. The hill has a radius of 14.6m. What is the skier's initial velocity?
(Problem continues but only need help for this part)

Homework Equations



K=1/2mv2
G=mgh
Fc=(mv2)/r

The Attempt at a Solution



So since the skier's "apparent weight" is 5/6 of his/her real weight, then force normal is = 5mg/6.
Since only gravity is holding the skier in a circle, then force centripetal = 5/6 force gravity.
([STRIKE]m[/STRIKE]v2)/r = (5/6)([STRIKE]m[/STRIKE]g)
v=sqrt(5/6)(9.81)(14.6)
v=10.92m/s
This is the speed of the skier at the top of the hill

Then we set the conservation of energy equation
Kinitial = Khill + G
K = 1/2mv2 + mgh = (0.5)(50.0)(10.922) + (50.0)(9.81)(4.00)
K = 2983.875J + 1962J = 4945.875J
1/2mv2 = 4945.875J
v= sqrt [(4945.875J)(2)/(50.0kg)]
v=14.07m/s


It seems logical to me but for some reason it is not correct : /
Help appreciated thanks :)
 
Physics news on Phys.org
Virtuozzo;3532884 [h2 said:
The Attempt at a Solution[/h2]

So since the skier's "apparent weight" is 5/6 of his/her real weight, then force normal is = 5mg/6.
Since only gravity is holding the skier in a circle, then force centripetal = 5/6 force gravity.

Draw the free body diagram and think over how is the centripetal force related to gravity and normal force.

ehild
 
Hi Virtuozzo and hi PF! I'm losing my virginity here :blushing:

You've got the right idea for the second (conservation of energy) part, only your velocity from the first part is wrong. For the first part, note that the skier isn't moving in the vertical direction, so this is a statics problem: The sum of all forces (in the y-direction) equals zero, and you've got three distinct forces at play as ehild mentioned, so your equation should have three terms. Hint: the top-of-the-hill velocity should be less than half of the value that you got.

Also, it's good practice to wait until the end of the problem to plug in your numbers to ensure accuracy. Good luck!
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top