Question

A block being pulled to the right by a 10N force acting 30° above the horizontal undergoes uniform motion on a level surface . The coefficient of sliding friction between the block and the surface is 0.4 what is the acceleration of the block

Answer 1

Answer on Question #65767 - Physics Mechanics Relativity

A block being pulled to the right by a 10N force acting $30{}^{\circ}$ above the horizontal undergoes uniform motion on a level surface. The coefficient of sliding friction between the block and the surface is 0.4 what is the acceleration of the block

Data:

$F = 10N$

$\alpha = 30{}^{\circ}$

$\mu = 0.4$

Solution:

According to $2^{\text{nd}}$ Newton's law: $\vec{F}_{sum} = m\vec{g} + \vec{F}_{fr} + \vec{N} + \vec{F} = m\vec{a}$

on $\mathrm{OX}: - F_{fr} + F*\cos \alpha = ma$

on OY: $mg = N + F * \sin \alpha$

As $F_{fr} = \mu N$ : $ma = F(\cos \alpha + \mu \sin \alpha) - \mu mg$ ; $\pmb{a} = \frac{F(\cos \alpha + \mu \sin \alpha) - \mu mg}{m}$ .

Acceleration of the block depends on its mass(no info about its mass)

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