A 65.0 kg skier slides down a 37.20 slope with mu = 0.107. What is the friction force?

Suppose you are navigating a spacecraft far from other objects. The mass of the spacecraft is 2.0 104 kg (about 20 tons). The ro

elena-14-01-66 [18.8K]

Answer:

r = <2.640 10⁶, 1.01 10⁸, 0> m

Explanation:

For this exercise we are going to solve it for each direction separately,

we locate a fixed reference frame in space at the height of the rocket, such that the position of the rocket is

r₀ = <4, 4, 0> 10³ m

X axis

the initial velocity of the ship on this axis is v₀ₓ = 0, when it passes through the point x₀ = 4 km it ignites the rockets, experiencing a force of Fₓ = 7.0 10⁵ N for 21 s and the rockets turn off

They ask where it is after one hour t = 1 h = 3600 s

Let's apply Newton's second law

Fₓ = m aₓ

aₓ = Fₓ / m

aₓ = 7.0 10⁵/2 10⁴

aₓ = 3.5 10¹ m / s

Let's use kinematics to find the distance

for the first t₁ = 21 s the movement is accelerated

x₁ = x₀ + v₀ t₁ + ½ aₓ t₁²

x₁ = x₀ + ½ aₓ t₁²

x₁ = 4000 + ½ 35 21² = 4000 + 7717,.5

x₁ = 11717.5 m

this instant has a speed of

vₓ = v₀ₓ + aₓ t

vₓ = aₓ t ₁

vₓ = 35 21

vₓ = 735 m / s

the rest of the time there is no acceleration so it is uniform motion at this speed

t₂ = 3600 - 21

t₂ = 3576 s

vₓ = x₂ / t₂

x₂ = vₓ t₂

x₂ = 735 3576

x₂ = 2,628 10⁶ m

the total distance traveled in this direction is

x_total = x₁ + x₂

x_total = 11717.5 + 2.628 10⁶

x_total = 2,640 10⁶ m

Y axis

on this axis it is in the initial position of y₀ = 4000 m, with an initial velocity of $v_{oy}$ = 28 10³ m / s and there is no force on this axis F_{y} = 0

The movement in this axis is uniform,

v_{y} = y / t

y = v_{y} t

y = 28 10³/3600

y = 1.01 10⁸ m

the total distance is

y_total = y₀ + y

y_total = 4000 + 1.01 10⁸

y_total = 1.01 10⁸ m

Z axis

the initial position is z₀ = 0, with an initial velocity of v₀ = 0 and in this axis there is no force F_{z} = 0

the movement is uniform

z = 0

the final position of the rocket after 1 h is

r = <2.640 10⁶, 1.01 10⁸, 0> m

7 0

2 years ago

A 17.6-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally

leonid [27]

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration a of

a = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ F = n - mg = 0

where ∑ F is the net force, n is the magnitude of the normal force, and mg is the weight of the block - it follows that n = mg ; and

• the net horizontal force is

∑ F = F - f = ma

where F is the applied force, f is kinetic friction, m is the block's mass, and a is the acceleration found earlier. F stretches the spring by x = 0.250 m, so we have

F - f = kx - µn = kx - µmg = ma

where k is the spring constant and µ is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ F = n - mg = 0

so that n = mg again; and

• the net horizontal force is

∑ F = F - f = 0

This time, F stretches the spring by y = 0.0544 m, so we have

F - f = ky - µmg = 0

Solve the equations in boldface for k and µ :

kx - µmg = ma

ky - µmg = 0

==> k (x - y) = ma

==> k = ma / (x - y)

==> k = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

ky - µmg = 0

==> µ = ky / (mg)

==> µ = (182 N/m) (0.0544 m) / ((17.6 kg) g) ≈ 0.0574

3 0

2 years ago

A SMALL cup of tea has water inside that is very hot, about 80 degrees Celsius. A LARGE bathtub has water inside it that is warm

Artist 52 [7]

Answer:

the water in the tea cup

Explanation:

Kinetic energy is the energy an object or particles possesses as a result of motion.

Liquid in higher temperature possesses more kinetic energy than liquid with lower temperature

7 0

2 years ago

A train 400 m long is moving on a straight track with a speed of 81.4 km/h. The engineer applies the brakes at a crossing, and l

Luda [366]

Answer:

86.5m

Explanation:

first convert km/h

then

81.4*1000/60*60=22.6

17.6*1000/60*60=4.89

then, x1/t1=x2/t2

we get

x2=400*4.89/22.6=86.5//

3 0

2 years ago

(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord ca

Aleks04 [339]

(a) Let $v$ be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

$a_{\rm rad} = \dfrac{v^2}R$

where $R$ is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

$F = (1.500\,\mathrm{kg}) a_{\rm rad}$

so that

$(1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N$

Solve for $v$ :

$v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}$

(b) The net force equation in part (a) leads us to the relation

$F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}$

so that $v$ is directly proportional to the square root of $R$ . As the radius $R$ increases, the maximum linear speed $v$ will also increase, so the cord is less likely to break if we keep up the same speed.

6 0

1 year ago

A 65.0 kg skier slides down a 37.20 slope with mu = 0.107.What is the friction force?