Ask question

Ask question

All categories

3 years ago

8

A tension force of 175 N inclined at 20.0° above the horizontal is used to pull a 40.0 - kg packing crate a distance of 6.00 m o

n a rough surface. If the crate moves at a constant speed, find (a) the work done by the tension force and (b) the coefficient of kinetic friction between the crate and surface.

1 answer:

Murljashka [212]3 years ago

4 0

Answer:

(a) Work done by the tension force is 987 J

(b) Coefficient of kinetic friction between the crate and surface is 0.495

Explanation:

(a)

Work done by any force F in moving an object by a distance d making an angle $\Theta$ with the direction of force is given by

$W=Fd\cos (\Theta )$

$W=175 \times 6 \times \cos (20 )J=987J$

Thus work done by the tension force is 987 J

(b)

Normal force on the crate is given by

$N= mg-F\sin \Theta =(40\times 9.8-175\times \sin 20)Newton$

=>N=332 Newtons

Since crate is moving with constant speed . Therefore using Newtons second law .

$Fcos(\Theta ) -\mu_k N=0$

Where $\mu_k$ =coefficient of kinetic friction

$\therefore \mu_k=\frac{F\cos (\Theta )}{N}$

=> $\mu_k=\frac{175\times \cos 20}{332}$

=> $\mu _k= 0.495$

Thus coefficient of kinetic friction between the crate and surface is 0.495

You might be interested in

What is the law of conservation of energy?

charle [14.2K]

The law of conservation of energy<span>, a fundamental concept of physics, states that the total amount of </span>energy<span> remains constant in an isolated system. It implies that </span>energy<span> can neither be created nor destroyed, but can be change from one form to another.</span>

6 0

2 years ago

A 120-kg roller coaster cart is being tested on a new track, and a crash-test dummy is loaded into itThe roller coaster starts f

Black_prince [1.1K]

Answer:

a) variation of the energy is equal to the work of the friction force

b) W = Em_{f} -Em₀ , c) he conservation of mechanical energy

Explanation:

a) In an analysis of this problem we can use the energy law, where at the moment the mechanical energy is started it is totally potential, and at the lowest point it is totally kinetic, we can suppose two possibilities, that the friction is zero and therefore by equalizing the energy we set the velocity at the lowest point.

Another case is if the friction is different from zero and in this case the variation of the energy is equal to the work of the friction force, in value it will be lower than in the calculations.

b) the calluses that he would use are to hinder the worker's friction force and energy

W = Em_{f} -Em₀

N d = ½ m v² - m g (y₂-y₁)

y₂-y₁ = 35 -10 = 25m

c) if there is no friction, the physical principle is the conservation of mechanical energy

If there is friction, the principle is that the non-conservative work is equal to the variation of the energy

7 0

3 years ago

70 POINTS AND BRAINLIEST PLEASE HELP!!!The Moon orbits Earth. This orbit causes the Moon to look different over the course of ab

Rudik [331]

Answer:

It always acurs after a 1st quarter

do you have photo?

Explanation:

8 0

3 years ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

3 0

1 year ago

21. (a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500-μC charge and flies due west at a

beks73 [17]

Answer:

Explanation:

The picture attached shows the full explanation and i hope it helps. Thank you

6 0

3 years ago