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morpeh [17]
3 years ago
7

A box is sitting on the ground and weighs 100 kg and the coefficient of friction is 0.23. Is it easier to push by applying the f

orce?

Physics
1 answer:
Hitman42 [59]3 years ago
7 0

Answer:

-No

Explanation:

Given that

Mass of box = 100 kg

Coefficient of friction ,μ= 0.23

We know that friction force depends on the normal force acts on the box

Fr= μ N

When we pull the box then :

Normal force N= mg

Friction force Fr=  μ mg                  

When we push the box :

Lets take pushing force = F

θ =Angle make by pushing force from the vertical line

Normal force

N = mg + F cosθ

Fr= μ ( mg + F cosθ )

The friction force is more when we push the box.That is why this is not easier to push the box.

Therefore answer is ---No

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The mass of the van will be 2.5 kg, Mass is found as the ratio of the force and the acceleration.

<h3>What is force?</h3>

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

F=ma

5.5 kN = m × 2.2m/s²

m= 2.5 kg

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A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has
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Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration (a) consist in two components, <em>radial</em> (a_{r}) and <em>tangential</em> (a_{t}), in meters per square second:

a_{r} = \omega^{2}\cdot r (1)

a_{t} = \alpha \cdot r (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

a = \sqrt{a_{r}^{2}+a_{t}^{2}} (3)

Where:

r - Radius of the wheel, in meters.

\omega - Angular speed, in radians per second.

\alpha - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

\omega = \omega_{o}+ \alpha\cdot t (4)

Where:

\omega_{o} - Initial angular speed, in radians per second.

t - Time, in seconds.

If we know that r = 0.1\,m, \alpha = 2\,\frac{rad}{s^{2}}, \omega_{o} = 0\,\frac{rad}{s} and t = 0.60\,s, then the total linear acceleration is:

\omega = \omega_{o}+ \alpha\cdot t

\omega = 1.2\,\frac{rad}{s}

a_{r} = \omega^{2}\cdot r

a_{r} = 0.144\,\frac{m}{s^{2}}

a_{t} = \alpha \cdot r

a_{t} = 0.2\,\frac{m}{s^{2}}

a = \sqrt{a_{r}^{2}+a_{t}^{2}}

a \approx 0.246\,\frac{m}{s^{2}}

The total linear acceleration is approximately 0.246 meters per square second.

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