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Vladimir79 [104]
3 years ago
12

Please help.: Sliding from left to right in a straight line on a horizontal steel surface, an aluminum block weighing 20 newtons

is acted on by a 2.4 newton friction force. The block will be brought to rest by the friction force in a distance of 10 meters. Determine the magnitude of the acceleration of the block as it is brought to rest by the friction force. (Show all work)
Physics
2 answers:
Vlad [161]3 years ago
7 0
Mass of the block = 2/g = 2/9.8 =0.204 kg. 

Frictional force = mass * acceleration 

a = F/ m where F is the frictional force = 2.4 

a = 2.4/ 0.204 = 11.76 m/s^2 
-BARSIC- [3]3 years ago
7 0

Answer:

-1.176 m/s²

Explanation:

Given:

weight of aluminium block is w = 20.0 N

frictional force, f = 2.4 N

final velocity, v = 0

distance covered, d = 10 m

From Newton's second equation of motion,

F = ma

where, m is the mass and a is the acceleration.

mass of the block, m g = 20.0 N

m = 2.04 kg

The block is brought to rest by friction force:

f =- m a

⇒ 2.4 N = -(2.04 kg)(a)

⇒a = -1.176 m/s²

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The 21 kg mass is attached by a cord to a mass hanging over the edge of the table. The frictional force between the mass and the
sasho [114]

Answer:

<em>147.3 N</em>

Explanation:

Two-Mass Systems

To solve a system where two masses are interacting with each other, we must set up the formulas by applying Newton's second law for each mass. Then, we find the required magnitudes by solving a system of equations.

Our system consists of a hanging object with mass m_2 attached to an object of mass m_1 lying in a table which applies a known friction force of F_R=121 N. We also know the system accelerates at 0.73\ m/sec^2. The situation is pictured in the image below.

Analyzing the forces acting upon mass m_1 we have, in the horizontal axis, where movement is taking place:

\displaystyle T-F_R=m_1\ a

Where T is the rope's tension force. Now taking the vertical axis of the second mass, we have

\displaystyle T-W_2=-m_2\ a

The acceleration is negative since it's directed downwards, contrary to the positive default direction (right and up). Subtracting both equations:

\displaystyle W_2-F_R=m_1\ a+m_2\ a

Solving for W_2

\displaystyle w_2=F_R+m_1\ a+m_2\ a

We know that

\displaystyle W_2=m_2\ g

so, the above formula becomes

\displaystyle m_2\ g=F_R+m_1\ a+m_2\ a

Rearranging and factoring

\displaystyle m_2(g-a)=F_R+m_1\ a

Solving for m_2

\displaystyle m_2=\frac{F_R+m_1\ a}{g-a}

Let's use our known data

\displaystyle m_2=\frac{121+21(0,73)}{9,8-0,73}=\frac{136.33}{9.07}

\displaystyle m_2=15.03\ kg

Finally, we compute the object's weight

\displaystyle W_2=m_2.g=15.03(9.8)=147.3\ N

3 0
3 years ago
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