Use the work-energy theorem to solve each of these problems. you can use newton’s laws to check your answers. (a) a skier moving at 5.00 m/s encounters a long, rough horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. how far does she travel on this patch before stopping?

d = 5.8 m

Explanation:

Principle of work and energy

ΔE = Wf

ΔE = Ef-Ei

ΔE : mechanical energy change

Wf : Work done by kinetic friction force

Ef : final mechanical energy

Ei : initial mechanical energy

K =(1/2 )mv² :  Kinetic energy

U= mgh   :Potential energy

m: mass (kg)

v : speed (m/s)

h: high (m)

Data

vi= 5.00 m/s

vf=0

hi=0

hf=0

μk=0.220

Kinetic friction force

fk = μk* FN

FN = W  : normal force (N)

W = m*g : weight  (N)

FN= 9.8*m (N)

fk =0.220*9.8*m

fk = (2.156)*m    Equation (1)

Work done by the  kinetic friction force

Wf = -fk*d  (J)  Equation (1)

d: distance traveled by force

Principle of work and energy to the skier

ΔE = Ef-Ei

Ef = Kf + Uf = 0

Ei = Kf + Uf = (1/2)(m)(5)² + 0

ΔE = -(1/2)(m)(5)²

ΔE = Wf

0- (1/2)(m)(5)² = -fk*d

We replace fk = (2.156)*(m )of the equation (1)

-(1/2)(m)(5)²= -(2.156)*(m)*d

We divide by (-m ) both sides of the equation

(1/2)(5)²=  (2.156)*d

12.5=  ( 2.156)*d

d = (12.5) / ( 2.156)

d = 5.8 m

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