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A 40KGskier skis directly down a frictionless slope angled at 100to the horizontal. Assume the skier moves in the negative direction of an xaxis along the slope. A wind force with component role="math" localid="1657169142011" Fxacts on the skier. (a) What is Fxif the magnitude of the skier’s velocity is constant, (b) What isif the magnitude of the skier’s velocity is increasing at a rate of 1.0m/s2, and (c) What is Fxif the magnitude of the skier’s velocity is increasing at a rate of2.0m/s2?

Short Answer

Expert verified

(a) The x component of wind force when skier moves with constant velocity is, fx=68N(uphill).

(b) The x component of wind force when skier moves with the acceleration of role="math" localid="1657169523821" a=1.0m/s2is,Fx=28N(uphill).

(c) The x component of wind force when skier moves with the acceleration of a=2.0m/s2is,Fx=-12N (downhill).

Step by step solution

01

Given data

  • The mass of the skier is, m=40kg.
  • The inclination angle with the horizontal is, θ=100.
  • The acceleration of a skier is, a=1.0m/s2.
  • The acceleration of the skier is,a=2.0m/s2.
02

Understanding the concept of Newton’s second law

As per Newton’s second law, the force acting on the object is directly proportional to the acceleration of the body. The force is equal to the product of mass and acceleration of the body. A free-body diagram shows the forces that are acting on the body along with their directions.

By using Newton’s second law x component of wind force when skier moves with constant velocity and with acceleration can be found.

Formulae:

F→net=ma→

Here,F→net is the net force, m is mass anda→ is acceleration.

03

Draw the free body diagram

04

a) Calculate Fx when skier moves with constant velocity

By applying Newton’s second law along x axis as,

Fnet→=ma→

The skier is moving with constant velocity hence it has zero acceleration. Consider the sign convention according to the direction of skier

Fx-mg.sinθ=0Fx=mg.sinθ=40kg×9.8m/s2×sin(100)=68N(uphill)

Hence, the value of role="math" localid="1657170901844" Fxwhen skier is moving with constant velocity is 68N .

05

b) Calculate Fx when skier moves with acceleration 1.0m/s2

The skier is going down with some acceleration, hence according to the Newton’s second law,

Fx-mg.sinθ=-maFx=mg.sinθ-ma=40kg×9.8m/s2×sin(100)-40kg×1.0m/s2=28N(uphill)

Hence, the value of Fxwhen skier is going down with 1.0m/s2 the acceleration is28N.

06

c) Calculate Fx when skier moves with acceleration 2.0m/s2

The skier rate is increasing in downward direction, hence according to the Newton’s second law,

Fx-mg.sinθ=-maFx=mg.sinθ-maFx=(40kg×9.8m/s2×sin(100))-(40kg×1.0m/s2)Fx=-12N(downhill)

Hence, the value of Fxwhen skier is going down with the 2.0m/s2acceleration is -12 N

The negative sign indicates that wind is also flowing in the downward direction, hence skier rate increases.

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