# IMAT 2014 Q53 [Acceleration]

A book of mass 0.40 kg rests on a horizontal surface with which it has a coefficient of dynamic friction of 0.50.

If this book is now pushed by an external horizontal force of 10 N, what will be its acceleration immediately after it has started to move?

[Assume the gravitational field strength is 10 Nkg­^{-1}, that air resistance is negligible and that the orientation of the book does not change.]

A. 25.0 ms^{-­2}
B. 12.5 ms^{-­2}
C. 50.0 ms^{-­2}
D. 20.0 ms^{-­2}
E. 15.0 ms^{-­2}

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The weight of the book, or the force pulling down on it, would be 4N.

This can be calculated by:

W=mg
W= 0.40 kg\times 10Nkg^{-1}= 4.0 N

The force of friction, which opposes the force of the new external horizontal force, can now we calculated using the coefficient of the mentioned dynamic friction.

The formula for dynamic friction is \mu= F/N, where:
\mu= coefficient
F= frictional force
N= normal force

Using this, we can say:

0.50= F/4
F= 0.50x4= 2N

Our frictional force is 2N. Our opposing external force is 10N. Our resultant force, or driving force of the book, is therefore 8N.

Now we must use newton’s first law, expressed by the formula:

F=ma

To find our acceleration.

8= 0.40a
a= 8/0.40= 20 ms^{-2}

Our answer is now clearly D.

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Can someone help with this question? Tried answering and got very confused - why do we need to get the weight if the question is asking about forces in the horizontal component?

I get that its acting on the book but I assumed the friction is acting opposite the ‘external horizontal force’?

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lad we need to calculate the frictional force for the net horizontal force. for calculating net horizontal force you need Normal force which acts opposite to weight of object, thus bringing it to vertical equilibrium. and thus frictional force = Coefficient * N = Coefficient * mg