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A 4kg table pushed to the right with an applied force of 50N. The table has a net acceleration of 10 m/s^2 to the right. What is the magnitude of the force of friction acting on the table?

Answer :

Answer:

Force of friction is 10 N.

Explanation:

Given:

Mass of the table is, [tex]m=4\textrm{ kg}[/tex]

Force acting towards right, [tex]F=50\textrm{ N}[/tex]

Net acceleration of the table, [tex]a=10\textrm{ }m/s^{2}[/tex]

Let the force of friction acting on the table to the left be [tex]f[/tex] N.

Therefore, net force acting on the table is given as,

[tex]F_{net}=F-f=50-f[/tex]

Now, as per Newton's second law of motion,

[tex]F_{net}=ma[/tex]

Equating the above two equations, we get

[tex]50-f=ma\\f=50-ma\\f=50-(4\times 10)\\f=50-40=10\textrm{ N}[/tex]

Therefore, the magnitude of the force of friction acting on the table is 10 N to the left direction.

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