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2 years ago

If the tension in the rope is 160 n, how much work does the rope do on the skier during a forward displacement of 270 m?

2 answers:

8 0

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on the skier is,

$W=T.S$

⇒ $W=270*160*cos\pi \\W=-43200 J$

Hence work done by the rope is - 43200 J.

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Mandarinka [93]2 years ago

6 0

The correct answer is - 43200 J.

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on the skier is,⇒

Hence work done by the rope is - 43200 J.

What is work ?

Work is the energy transferred to or from an object by applying force along a displacement in physics. It is frequently represented as the product of force and displacement in its most basic form. When a force is applied, it is said to do positive work if it has a component in the direction of the point of application's displacement. If a force has a component that is opposite the direction of displacement at the force's application point, it does negative work.

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What is the atomic number of an atom?

Sever21 [200]

Answer:

B, the number of protons.

Explanation:

It is the number of protons found in the nucleus of the atom.

4 0

2 years ago

A stubborn, 100 kgkg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around

jolli1 [7]

Answer:

No, the farmer is not able to move the mule.

Explanation:

Mass =100 kg

Force=F=800 N

The coefficient between the mule and the ground= $\mu_s=0.8$

$\mu_k=0.5$

Static friction force,f= $\mu N$

Normal force=N=mg

Static friction force,f= $\mu_s mg=0.8\times 100\times 9.8=784 N$

Using $g=9.8m/s^2$

F<f

Static friction force is greater than applied force.

Therefore , the farmer is not able to move the mule.

4 0

3 years ago

Determine the amount of work done by the applied force when a 87 N force is applied to move a 15 kg object a horizontal

elena-s [515]

Answer:

391.5 J

Explanation:

The amount of work done can be calculated using the formula:

W = F║d
where the force is parallel to the displacement

Looking at the formula, we can see that the mass of the object does not affect the work done on it.

Substitute the force applied and the displacement of the object into the equation.

W = (87 N)(4.5 m)
W = 391.5 J

The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.

5 0

2 years ago

Read 2 more answers

A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

$F_k = F_{W,E}$

$kx_1 = mg$

The extension of the spring due to the weight of the object on Moon is a value of $x_2$ , then

$kx_2 = mg_m$

Recall that gravity on the moon is a sixth of Earth's gravity.

$kx_2 = m\frac{g}{6}$

$kx_2 = \frac{1}{6} mg$

$kx_2 = \frac{1}{6} kx_1$

$x_2 = \frac{1}{6} x_1$

We have that the displacement at the earth was $x_1 = 0.3m$ , then

$x_2 = \frac{1}{6} 0.3$

$x_2 = 0.05m$

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0

2 years ago

A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is hold

lubasha [3.4K]

Answer:

to overcome the out of friction we must increase the angle of the plane

Explanation:

To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.

X axis

fr - Wₓ = m a (1)

Y axis

N- $W_{y}$ = 0

N = W_{y}

let's use trigonometry to find the components of the weight

sin θ = Wₓ / W

cos θ = W_{y} / W

Wₓ = W sin θ

W_{y} = W cos θ

the friction force has the formula

fr = μ N

fr = μ Wy

fr = μ mg cos θ

from equation 1

at the point where the force equals the maximum friction force

in this case the block is still still so a = 0

F = fr

F = (μ mg) cos θ

We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.

This is the force that balances the friction force, any force slightly greater than F initiates the movement.

Consequently, to overcome the out of friction we must increase the angle of the plane

the correct answer is to increase the angle of the plane

4 0

3 years ago