Respuesta :
Answer:
Explanation:
force of friction = μ mg
= .3 x 62 x 9.8
= 182.28 N
Work done by friction = 182.28 x 4.8
= 874.94 J
Energy loss by skier = 874.94 J
Kinetic energy after he passes the rough patch
= 1/2 x 62 x 6.8² - 874.94
= 1433.44 - 874.94
= 558.5 J
Velocity after crossing the patch v
1/2 x 62 x v² = 558.5
v² = 18
v = 4.2426 m /s
velocity at the bottom
v² = u² + 2gh
= 18 + 2 x 9.8 x 2.5
= 67
v = 8.18 m /s
b ) internal energy generated =
Work done by friction . = 874.94 J
Answer:
a) [tex]v_f=8.186\ m.s^{-1}[/tex]
b) [tex]E=874.944\ J[/tex]
Explanation:
Given:
- mass of skier, [tex]m=62\ kg[/tex]
- velocity of skier on frictionless surface, [tex]u=6.8\ m.s^{-1}[/tex]
- length of rough patch, [tex]l=4.8\ m[/tex]
- coefficient of friction between the patch and the skis, [tex]\mu=0.3[/tex]
- height of the down-hill, [tex]h=2.5\ m[/tex]
a)
Frictional force acting on the rough patch while motion:
[tex]f=\mu.N[/tex]
where:
[tex]N=[/tex] reaction normal to the surface acting on the body
[tex]f=0.3\times (62\times 9.8)[/tex]
[tex]f=182.28\ N[/tex]
Now this force will cause deceleration in the speed if the skier and is given as:
[tex]d=\frac{f}{m}[/tex]
[tex]d=\frac{182.28}{62}[/tex]
[tex]d=2.94\ m.s^{-2}[/tex]
Using equation of motion:
[tex]v^2=u^2-2d.l[/tex]
[tex]v^2=6.8^2-2\times2.94\times 4.8[/tex]
[tex]v=4.24\ m.s^{-1}[/tex]
Now the final velocity at the bottom of the hill:
[tex]v_f^2=v^2+2g.h[/tex]
[tex]v_f^2=18.016+2\times 9.8\times 2.5[/tex]
[tex]v_f=8.186\ m.s^{-1}[/tex]
b)
Internal energy is generated due to the kinetic frictional force and the distance covered on the patch:
(Using work-energy equivalence)
[tex]E=f.l[/tex]
[tex]E=182.28\times 4.8[/tex]
[tex]E=874.944\ J[/tex]
