Question

A 62.0 kgkg skier is moving at 6.90 m/sm/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 mm long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 mm high.How fast is the skier moving when she gets to the bottom of the hill?
How much internal energy was generated in crossing the rough patch?

Answer 1

Answer:

the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.

the internal energy generated in crossing the rough patch is 820.26 J

Explanation:

Given that,

Mass, m = 62 kg,

Initial speed,  = 6.90 m/s

Length of rough patch, L = 4.50 m,

coefficient of friction,  = 0.3

Height of inclined plane, h = 2.50 m

$g=9.8 m/s^2$ is the acceleration due to gravity

According to energy conservation equation,

Part (b)

The internal energy generated when crossing the rough patch is equal (in magnitude) to the work done by the friction force on the skier.

The magnitude of the friction force is:

$F_f=\mu mg$

Therefore, the work done by friction is:

$W=-F_f d =-\mu mg d$

The negative sign is due to the fact that the friction force is opposite to the direction of motion of the skier

Substituting,

$W=-(0.300)(62.0)(9.8)(4.50)=-820.26 J$

So, the internal energy generated in crossing the rough patch is 820.26 J

Part (a)

If we take the bottom of the hill as reference level, the initial mechanical energy of the skier is sum of his kinetic energy + potential energy:

$E=K_i +U_i = \frac{1}{2}mu^2 + mgh$

After crossing the rough patch, the new mechanical energy is

$E'=E+W$

where

W = -820.26 J is the work done by friction

At the bottom of the hill, the final energy is just kinetic energy,

$E' = K_f = \frac{1}{2}mv^2$

where v is the final speed.

According to the law of conservation of energy, we can write:

$E+W=E'$

So we find v:

$\frac{1}{2}mu^2 +mgh+W = \frac{1}{2}mv^2\\\\v=\sqrt{u^2+2gh+\frac{2W}{m}}\\\\=\sqrt{6.9^2+2(9.8)(2.50)+\frac{2(-820.26)}{62.0}}\\\\=8.38 m/s$

Thus, the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.

Answer 2

The question repeated the units attached the the values

So in order of their appearance in the question, the correct units are;

62kg, 6.9m/s, 4.5m, 0.30, 2.5m

Answer:

A) Speed at bottom of hill = 8.38 m/s

B) Internal energy generated in crossing the patch = 820.26J

Explanation:

From the question, it's clear that there are 2 distinct stages of of her skis. The first stage is when she is on the rough patch while the second stage is when she skis down the hill.

A) For the first stage when she is on the rough patch;

If we apply Newton's second law of motion to the skis along the vertical direction, we obtain;

ΣF_y = N - mg = 0

Thus,N = mg

m = 62kg,thus,N = 62 x 9.8 = 607.6N

Now, let's find the kinetic friction. It's given by; f_k = μN

Where μ is coefficient of kinetic friction

Thus, f_k = 0.3 x 607.6 = 182.28N

Now, let's calculate the work done by this frictional force,

So, W_fk = f_k x distance = 182.28 x

4.5 = 820.26J

This work is done in a direction opposite to the displacement and it will have a negative sign. Thus,

W_fk = - 820.26J

Now, since the skier skies horizontally and perpendicular to the gravitational force, the work done due to gravity is zero. Thus,

W_grav = 0

Let's calculate the kinetic energy at the beginning of the rough patch.

K1 = (1/2)m(v1)²

K1 = (1/2) x 62 x (6.9)²= 1475.91J

Also,let's calculate the kinetic energy at end of rough patch.

K2 = (1/2)m(v2)²

where v2 is final velocity at end of rough patch

K2 = (1/2)(62)(v2)² = 31(v2)²

Now, the total work done when other forces other than gravity do work, is given by ;

W_total = W_other + W_grav = K2 - K1

In this case, W_other is W_fk

Thus,

- 820.26J + 0 = 31(v2)² - 1475.91J

31(v2)² = 1475.91J - 820.26J

31(v2)² = 655.65

(v2)² = 655.65/31

v2 =√21.15 = 4.6 m/s

Now, for the second stage when she skis down the hill;

In this case the only force acting is gravity, thus, W_other = 0

Work done by gravity = mgh

W_grav = 62 x 9.8 x 2.5 = 1519 J

Now K2 will be kinetic energy at top of hill while K3 will be kinetic energy at bottom of hill.

Thus,

K2 = (1/2)m(v2)²

K2 = (1/2)(62)(4.6)²

K2 = 655.96J

Similarly,

K3 = (1/2)m(v3)²

Where v3 is velocity at bottom of hill.

Thus,

K3 = (1/2)(62)(v3)²

K3 = 31(v3)²

Again, the total work done when other forces other than gravity do work, is given by ;

W_total = W_other + W_grav = K3 - K1

So,

W_other is zero.

Thus,

1519J = 31(v3)² - 655.96J

31(v3)² = 1519J + 655.96J

31(v3)² = 2174.96J

(v3)² = 2174.96/31

v3 = √70.16

v3 = 8.38 m/s

B) Workdone by non-conservative forces manifests itself as changes in the internal energy of bodies. Now, frictional force experienced in the first stage in crossing the patch is a non - conservative force. Thus,

Internal energy = W_fk = 820.26J