Draw free body diagramm of all forces acting on skier

AI Thread Summary
The discussion focuses on analyzing the forces acting on a skier descending a frictionless slope, specifically addressing the effects of air resistance modeled as a drag force proportional to velocity. A free body diagram is drawn to illustrate the forces, with clarification on whether the wind direction is parallel to the slope or horizontal. The participants derive a differential equation for the skier's velocity, noting that the acceleration cannot be treated as constant due to the velocity-dependent drag. The solution for the skier's velocity as a function of time is identified, leading to the expression for terminal velocity. Overall, the conversation emphasizes the relationship between forces, acceleration, and velocity in the context of the skier's motion.
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Homework Statement



A skier of mass m is skiing down a frictionless slope. the skier starts form rest at t = 0 and is subject to a velocity dependant drag force due to air resistance of the form F = -bv, where b is a constant.

a)draw free body diagramm of all forces acting on skier

b) wire a differential equation that can be used to solve for the velocity of the skier as a function of time

c) find expression for Terminal velocity

d) solve b and ditermine the velocity of the skier as a function of time

Homework Equations





The Attempt at a Solution



a) note the attached diagramm. does the the wind come parallel to the hill of horizontally

b)

Fx = ma = mgsin\theta - bv

v = (mgsin\theta - ma)/b

i believe i need to find an equation for a which involvs t, do i just use one of the kinematic equations

the answer is v(t) = (mg sinθ / b) (1 – e-bt/m) where did the who e part come from
 
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joemama69 said:

The Attempt at a Solution



a) note the attached diagramm. does the the wind come parallel to the hill of horizontally
Since the air resistance force is -bv, it is directly opposite to the direction of the skier's velocity. Is the skier's velocity parallel to the hill or horizontal?
The attachment seems to be missing.
b)

Fx = ma = mgsin\theta - bv

v = (mgsin\theta - ma)/b

i believe i need to find an equation for a which involvs t, do i just use one of the kinematic equations

the answer is v(t) = (mg sinθ / b) (1 – e-bt/m) where did the who e part come from
It cannot come from the kinematic equations, which only work when acceleration is constant.

It comes from the equation you wrote in part (b),

ma = mg sinθ - bv​

Hint: what equation defines a in terms of v?
 


Sorry, here's the attachment

the velocity is parallel to the ground.

What do you mean by it came from the equation from part b, i can't plug the inquation into itself
 

Attachments



Your free body diagram looks good.

joemama69 said:
the velocity is parallel to the ground.
Correct.

What do you mean by it came from the equation from part b, i can't plug the inquation into itself
True, lol. No, you need to substitute another expression in for a first. Refer to my hint:
Redbelly98 said:
Hint: what equation defines a in terms of v?
 


a = dv/dt
 
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