Help Solve Skier Problem Homework

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The skier problem involves calculating the final velocity of a skier descending a 20m hill with an initial velocity of 5 m/s and a weight of 50kg, while considering a frictional force of 40N acting uphill. The total energy at the start includes both kinetic and potential energy, calculated to be 10425J. To find the final velocity, the work done by the friction force must be determined and subtracted from the total energy. The work-energy theorem can then be applied to find the skier's velocity at the bottom of the hill. Proper understanding of energy conservation and the impact of friction is crucial for solving the problem.
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Homework Statement


if a skier goes down a hill that is 20m high and 70 m long perpendicular to the height... and she weighs 50kg and her intial veloctiy is 5 m/s. how fast is she going at the bottom if the force of gravity up the hill is 40N??
I just don't get it!
PLease help me


Homework Equations


total energy= kinetic energy + gravitational energy

work = Fd



The Attempt at a Solution


Et= .5x50x(5x5)+ 50 x 9.8 x20
= 10425

and i don't know where to go from here please help me!
i have a test tomorrow and i don't understand
 
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Welcome to PF!

Hi peyton! Welcome to PF! :smile:
peyton said:
Homework Statement [/b]
if a skier goes down a hill that is 20m high and 70 m long perpendicular to the height... and she weighs 50kg and her intial veloctiy is 5 m/s. how fast is she going at the bottom if the force of gravity up the hill is 40N??
I just don't get it!

What a terrible question :frown: …

how can gravity be up the hill … doesn't your professor know what "up" means??

and how can anyone have a weight in kg? weight is in Newtons, mass is in kg :rolleyes:

hmm … if her mass is 50kg, the force of gravity down the hill would be about 140 N, not 40 N (it isn't, is it?) …

i've no idea what this question means …

if you have to answer it, I suggest you ignore the 40 N, which is indeed what you've already done …
Et= .5x50x(5x5)+ 50 x 9.8 x20
= 10425

… now what speed does that correspond to? :smile:
total energy= kinetic energy + gravitational energy


"total energy" isn't really a helpful concept (and nobody calls it "Et" :wink:) …

I suggest you say KE + PE = constant, or ∆(KE) = -∆(PE)
 
btw i am Canadian
the 10425J
and it was the force of friction was 40N up hill

i need to find the velocity at the bottom of the hill
 
peyton said:
and it was the force of friction was 40N up hill

D'oh! :rolleyes:

In that case, you need to find the work done by the friction force, and use the work-energy theorem.
btw i am Canadian

ahh! :redface:
 
then what ?
find the difference then use that force to find the velocity at the bottom of the slope
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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