Chapter 0S, Problem 055 Two blocks are in contact on a frictionless table. A hor

Question

Chapter 0S, Problem 055 Two blocks are in contact on a frictionless table. A horizontal force is applied to the larger block, as shown in the figure. a) If m1-2.8 kg m2 = 1.7 kg, and F 2.9 N find the magnitude of the force between the two blocks. (b) Assume instead that a force of the same magnitude F is applied to the smaller block but in the opposite direction, and calculate the magnitude of the force between the blocks. (Why is the value calculated in (b) not the same as that calculated in (a)?) (a) Number Units (b) Number Units

Explanation / Answer

Let the mass of first block be M = 2.8 kg

Mass of the second block be m = 1.7 kg

Let the applied horizontal force be F = 2.9 N

Now the key point in solving the problem is that the acceleration of the two blocks is same as they are in contact.

Let this acceleration be a

a) From Newton's second law , net force is equal to product of mass and acceleration.

Here no friction is acting , so applied force can be taken as net force for the " system of two blocks " ( not for the single blocks considered at a time )

Now total mass of two block system be M + m

So net force F = ( M + m ) a

So we can write a = F / ( M + m )

= 2.9 / ( 2.8 + 1.7 )

= 2.9 / 4.5

= 0.6444 m/s2

Now the net force on the second block m is f = m a

= 1.7 × 0.6444

= 1.095 N

The same force is applied by second block on first block as a reaction.

So force between two blocks is f = 1.095 N

b) Similarly if a force is applied on second block in opposite direction , we can solve in the same way

Since the magnitude of the applied force has not changed , acceleration will be same.

So acceleration of the two blocks is a = 0.6444 m/s2

Now the net force on the first block is F1 = M a

= 2.8 × 0.6444

= 1.8044 N

This would be the force between two blocks.

Now a question arises why the force is different when acceleration is same.

The answer lies in our calculation only.

First the block of larger mass pushes a lighter block , so a small force is enough.

In the second case , larger block is to be pushed by the smaller one. So a larger force is required.

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