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snow_lady [41]
3 years ago
11

A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.30 m/s . The coefIficient o

f kinetic friction between the box and the surface is 0.25.
Physics
1 answer:
Artemon [7]3 years ago
8 0

Answer:

Explanation:

Mass =11.2kg

Constant velocity =3.3m/s

μk=0.25

Since the body is moving in constant velocity, then the acceleration is zero(0).

ΣF = Σ(ma)

The normal force acting on the body is upward and the weight is acting downward

Then ΣFy=0

Therefore, N=W

W=mg=11.2×9.8=109.76N

So, N=W=109.76N

Frictional force is given as

Fr=μkN

Fr=0.25×109.76

Fr=27.44N

Frictional force acting against the motion is 27.44N

Then the forward force moving the body forward

ΣF = Σ(ma)

Since a = 0

Then,

ΣF = 0

F-Fr=0

Then F=Fr

So the force moving the body forward is 27.44N

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How do I differentiate between final and initial velocity in Physics?​
pychu [463]

Answer:

Initial velocity describes how fast an object travels when gravity first applies force on the object. On the other hand, the final velocity is a vector quantity that measures the speed and direction of a moving body after it has reached its maximum acceleration.

Explanation:

3 0
3 years ago
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
3 years ago
What is the difference between kinetic and potential energy and how do they work?
Iteru [2.4K]

To explain, I will use the equations for kinetic and potential energy:

PE = mgh\\KE = \frac{1}{2}mv^{2}

<h3>Potential energy </h3>

Potential energy is the potential an object has to move due to gravity.  An object can only have potential energy if 1) <u>gravity is present</u> and 2) <u>it is above the ground at height h</u>.  If gravity = 0 or height = 0, there is no potential energy.  Example:

An object of 5 kg is sitting on a table 5 meters above the ground on earth (g = 9.8 m/s^2).  What is the object's gravitational potential energy?  <u>(answer: 5*5*9.8 = 245 J</u>)

(gravitational potential energy is potential energy)

<h3>Kinetic energy</h3>

Kinetic energy is the energy of an object has while in motion.  An object can only have kinetic energy if the object has a non-zero velocity (it is moving and not stationary).  An example:

An object of 5 kg is moving at 5 m/s.  What is the object's kinetic energy?  (<u>answer: 5*5 = 25 J</u>)

<h3>Kinetic and Potential Energy</h3>

Sometimes, an object can have both kinetic and potential energy.  If an object is moving (kinetic energy) and is above the ground (potential), it will have both.  To find the total (mechanical) energy, you can add the kinetic and potential energies together.  An example:

An object of 5 kg is moving on a 5 meter table at 10 m/s.  What is the objects mechanical (total) energy?  (<u>answer: KE = .5(5)(10^2) = 250 J; PE = (5)(9.8)(5) = 245 J; total: 245 + 250 = 495 J</u>)

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nata0808 [166]

Answer:

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Explanation:

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Answer:

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Explanation:

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