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kompoz [17]
3 years ago
15

Atwood machine. A 20kg box and a 30kg crate are attached to the two ends of a massless string that passes over a massless fricti

onless pulley. The system is released from rest. Without performing a calculation, rate, smallest to largest, the magnitudes of the tension in the string and the weights of box and crate, respectively. Calculate the acceleration of the system and the tension in the string
Physics
1 answer:
Verdich [7]3 years ago
5 0

Answer with Explanation:

We are given that

m_1=20 kg

m_2=30 kg

We have to find the acceleration of the system and the tension in the string.

30g-T=30a...(1)

T-20g=20a

T=20a+20g...(2)

Using equation (2) in equation (1)

30g-(20a+20g)=30a

30g-20a-20g=30a

10g=30a+20a=50a

a=\frac{10g}{50}=\frac{10\times 9.8}{50}=1.96 m/s^2

Where g=9.8m/s^2

Using the value of a in equation (2)

T=20(a+g)=20(1.96+9.8)=235.2 N

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One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what
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The coefficient of kinetic friction is 0.13

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Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

F_x=ma_x (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

n=7406.55 N (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

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4 years ago
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