Question

A 62.0 kg skier is moving at 6.80 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.80 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to the friction-free snow, she skis down an icy, frictionless hill 2.50 m high.(a) How fast is the skier moving when she gets to the bottom of the hill?(b) How much internal energy was generated in crossing the rough patch?

Answer 1

Answer:

a) $v_f=8.186\ m.s^{-1}$

b) $E=874.944\ J$

Explanation:

Given:

• mass of skier, $m=62\ kg$

• velocity of skier on frictionless surface, $u=6.8\ m.s^{-1}$

• length of rough patch, $l=4.8\ m$

• coefficient of friction between the patch and the skis, $\mu=0.3$

• height of the down-hill, $h=2.5\ m$

a)

Frictional force acting on the rough patch while motion:

$f=\mu.N$

where:

$N=$ reaction normal to the surface acting on the body

$f=0.3\times (62\times 9.8)$

$f=182.28\ N$

Now this force will cause deceleration in the speed if the skier and is given as:

$d=\frac{f}{m}$

$d=\frac{182.28}{62}$

$d=2.94\ m.s^{-2}$

Using equation of motion:

$v^2=u^2-2d.l$

$v^2=6.8^2-2\times2.94\times 4.8$

$v=4.24\ m.s^{-1}$

Now the final velocity at the bottom of the hill:

$v_f^2=v^2+2g.h$

$v_f^2=18.016+2\times 9.8\times 2.5$

$v_f=8.186\ m.s^{-1}$

b)

Internal energy is  generated due to the kinetic frictional force and the distance covered on the patch:

(Using work-energy equivalence)

$E=f.l$

$E=182.28\times 4.8$

$E=874.944\ J$

Answer 2

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