A skier, of unknown mass, is initially traveling towards a steep slope as shown

Question

A skier, of unknown mass, is initially traveling towards a steep slope as shown in picture below. Ignore air resistance.

(a) If the slope is smooth then calculate the minimum velocity magnitude the skier needs to have in order to barely reach the top of the slope.

(b) If the slope has a coefficient of kinetic friction of 0.22, then calculate the minimum velocity magnitude the skier needs to have in order to barely reach the top of the slope.

(c) If the slope has a coefficient of kinetic friction of 0.22, then calculate the minimum velocity magnitude the skier needs to have in order to reach the top of the slope with a velocity magnitude of 3.5 m/s.

Explanation / Answer

let,

height h=2.5 m

angle , theta=35 degrees

mass =m

initial velocity is vi

a)

by using conservation of energy,

1/2*m*vi^2=m*g*h

==> vi=sqrt(2*g*h)

vi=sqrt(2*9.8*2.5)

vi=7 m/sec

b)

coefficient of kinetic friction uk=0.22

frictional force fk=uk*mg*cos(theta)

distance travelled along slope is s=h/sin(theta)

by using conservation of energy,

1/2*m*vi^2=fk*s+m*g*h

1/2*m*vi^2=uk(mg)*cos(thtea)*h/sin(theta)+m*g*h

1/2*vi^2=(uk*g*cot(thtea)*h+g*h

vi=sqrt((1+uk*cot(thtea)*2*g*h)

vi=sqrt((1+0.22*cot(35))*2*9.8*2.5)

vi=8.025 m/sec

c)

vf=3.5 m/sec

by using conservation of energy,

1/2*m*vi^2=fk*s+m*g*h+1/2*m*vf^2

1/2*m*vi^2=uk(mg)*cos(thtea)*h/sin(theta)+m*g*h+1/2*m*vf^2

1/2*vi^2=(uk*cot(thtea)*g*h+g*h+1/2*vf^2

vi=sqrt(((1+uk*cot(thtea)*2*g*h)+vf^2)

vi=sqrt(((1+0.22*cot(35))*2*9.8*2.5)+3.5^2)

vi=8.75 m/sec

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