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Llana [10]
3 years ago
15

A car is being towed at a constant velocity on a horizontal road using a horizontal chain. The tension in the chain must be equa

l to the weight of the car in order to maintain a constant velocity.
True or False?
Physics
1 answer:
lukranit [14]3 years ago
3 0

Answer:

<em>The statement is false</em>

Explanation:

<u>Dynamics</u>

To analyze the situation stated in the question, we must apply some basic knowledge about dynamics, specifically Newton's laws that explain how acceleration is produced or not, depending on the forces acting on a system of particles.

The second Newton's law states the acceleration of an object of mass m can be found by knowing the net force Fn applied to it with the formula

F_n=m.a

Considering the kinetics equations, we know that

\displaystyle a=\frac{v_f-v_o}{t}

Where vf-vo is the change of velocity in time t. If an object has zero acceleration, it means its velocity doesn't change, it has constant velocity.

Under such conditions, then the net force is also zero.

We know the car is being towed horizontally at a constant velocity, meaning it has a net force equal to zero in the horizontal direction. In other words, all the horizontal forces are balanced.

We also know the chain is applying a horizontal force to tow the car, so there must be another force opposing tho that force making the car moving at a constant velocity. Although it's not mentioned, the friction force must be acting to stop the car and compensating the tension T exerted by the chain.

The friction force is given by

F_r=\mu N

Where \mu is the friction coefficient and N is the normal force, which is equal to the weight of the object W at the conditions of the problem.

Knowing the net force is zero, then

\mu W=T

If the tension is equal to the weight of the car, then

\mu T=T

simplifying

\mu=1

Thus, the only way the tension and the weight are equal is when the friction coefficient is 1, so the initial assumption is false

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Answer:

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Explanation:

Given that,

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Using formula of conservation of energy between point A and B

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U_{A}+0-fd=mgy+\dfrac{1}{2}mv^2

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Put the value into the formula

U_{A}=0.50\times1.60\times9.8\cos30\times6.55+1.60\times9.8\times6.55\sin30+\dfrac{1}{2}\times1.60\times(7.50)^2

U_{A}=88.8\ J

Hence, The amount of potential energy that was initially stored in the spring is 88.8 J.

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