All categories

2 years ago

Work is best described as?

Physics

1 answer:

lesya692 [45]2 years ago

5 0

Answer:

Explanation:

In physics, work can be described as force multiplied by the displacement of the object (distance caused by the force acted on the object).

Work is a form of energy, therefore, it is measured in joules.

If Alex pushed on an object with 10N of force, and the object moved 2 meters, the work of Alex will be equal to:

W = Ft

W = 10N * 2m

W = 20 Joules

But, if we have John, that pushed on an object with 2N of force, but, since the object was way smaller, therefore it had way less inertia, the object moved a distance equal to 10 meters. Let's also calculate his work.

W = Ft

W = 2N * 10m

W = 20 Joules

And we got the same result.

Work can also be used to calculate power.

Power is equal to P = Work / time

It can also be written as P = delta Work / delta time

Power is measured in joules per second.

With power you can make the difference between someone that got a 2 kg object up in 2 second, and someone that got a 2 kg object up in 5 second.

The more powerful one would be the one who managed to make the same work in less time.

Hope it Helped!

You might be interested in

4<br> Explica las características principales de las fuer-<br> zas de acción y reacción.

Naya [18.7K]

-Surgen de una interacción.
-Nunca aparece una sola: son dos y simultáneas.
-Actúan sobre cuerpos diferentes: una en cada cuerpo.
-Nunca forman un par de fuerzas: tienen la misma línea de acción.
-Un cuerpo que experimenta una única interacción no está en equilibrio, pues sobre el aparece una fuerza unica que lo acelera. Para estar en equilibrio se requieren por lo menos dos interacciones.

Las mas importantes son la 2,3,4 característica

5 0

3 years ago

John walks 1.00 km north, then turns right and walks 1.00 km east. His speed is 1.50 m/s during the entire stroll.a) What is the

avanturin [10]

Answer:

(a) 1.414 km

(b) 1.06 m/s

Explanation:

(a) For John:

Distance = 1 km north and then 1 km east

Speed = 1.5 m/s

total distance traveled = 1 + 1 = 2 km = 2000 m

Time taken to travel = Distance / speed

t = 2000 / 1.5 = 1333.3 seconds

Displacement = $\sqrt{1^{2}+1^{2}}=1.414 km$

(b) For jane :

Time is same as john = 1333.33 second

Distance = 1.414 km = 1414 m

Speed = distance / time = 1414 / 1333.33 = 1.06 m/s

3 0

3 years ago

The velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a veloci

nasty-shy [4]

Answer:

$\theta = 25.3^\circ$

Explanation:

The acceleration of the block can be found by the kinematics equations:

$v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2$

Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

$F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ$

7 0

4 years ago

A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the

andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

$F = mg$

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

$\sum \tau = 0$

Regarding the forces we have,

$3Mg-1mg=0$

Re-arrange to find M,

$M = \frac{m}{3}$

$M = \frac{50}{3}$

$M = 16.67Kg$

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0

3 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$

Hence, v0 = 6.86 m/s

4 0

3 years ago