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A 2.5-kg object moves across a rough horizontal surface. A force (F = 6.0 N) acts...

A 2.5-kg object moves across a rough horizontal surface. A force (F = 6.0 N) acts on the object as shown. The magnitude of the object’s acceleration is 1.2 m/s2. What is the magnitude of the force of friction acting on the block? The angle for the diagram is 30 degrees.

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Answer #1

Using Force balance on the diagram: (I'm assuming force F is applied above the horizontal)

Using force balancein horizontal direction:

F_net = F*cos - Ff

Ff = Friction force = ?

From Newton's 2nd law: F_net = m*a

F = Force applied = 6.0 N

= 30 deg

So,

Ff = F*cos - m*a

Ff = 6.0*cos 30 deg - 2.5*1.2

Ff = 2.2 N = Magnitude of friction force

Let me know if you've any query.

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