Question

A particle of mass 5kg is pulled along a rough horizontal surface by a string which is inclined at 60° to the horizontal. if the acceleration of the particle is 1/3×g ms^-2 and the coefficient of friction between the particle and the plane is 2/3, find the tension in the string​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

A particle of mass 5kg is pulled along a smooth horizontal surface by a horizontal string. The acceleration of the particle is 10$ms^{-2}$. The tension in the string is 20N.

Explanation:

Force balance in the horizontal direction hence force is equal to the string or the tension of the Horizontal side,

Again,

$T-m \times g \times sin \theta +\mu \times m \times g \times cos \theta =m \times a$

Where, T is the tension

m is the mass of the particle which is given in the equation as 5 kg.

g is the gravitational force

and $\theta$ is the angle i.e., 60°.

a is the acceleration.

simplify the equation to determine the value of tension,

$T=m \times g \times sin \theta -\mu \times m \times g \times cos \theta +m \times a$

Now, substitute the given data in the following equation,

$T=m \times (g \times sin \theta -\mu \times g \times cos \theta + a)$

$T=5 \times (g \times \frac{\sqrt{3}}{2} -\frac{2}{3} \times g \times \frac{1}{2} + \frac{g}{3})$

$T=5 \times (9.8 \times \frac{\sqrt{3}}{2} -\frac{1}{3} \times 9.8 + \frac{9.8}{3})$

Solving the equation we get the value of the tension of the string is 20 N.

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