Question

A 4.0- kg wooden block rests on a level table. The coefficient of friction between the block and the table is 0.23. A 5.0- kg mass is attached to the block by a horizontal string passed over a frictionless pulley of negligible mass. Now, the 5.0- kg mass is released and the whole system accelerates. 

What is the acceleration of the wooden block? 

What is the tension in the string during the acceleration? 

Determine the acceleration for the above situation when the coefficient of friction between the block and the table is 0.30. 

What is the tension when the coefficient of friction between the block and the table is 0.30? 

Answer 1

If coefficient of friction is 0.23:

Vertical forces on the 5kg block include:

1.gravitational force:mg downwards, where m is mass of block=5 kg and g is gravitational acceleration=9.8 m/s²

So,gravitational force=5*9.8=49 N downwards

2.Force due to tension in the string=T newtons upward.

Using newton's 2nd law:force=mass*acceleration,we get, 49-T=5a........................equation 1 where a is the acceleration of 5kg block,downward.

Since, the 4 kg block is not accelerating vertically, vertical forces must balance=>normal reaction N=weight of the block=mg (where m is mass of the block,g is gravitational acceleration)= 4*9.8=39.2 N

Horizontal forces on the 4 kg block include:

1.friction=kN,where k is coefficient of friction , N is normal reaction

So,friction=0.23*39.2=9.016 N towards left.

2.Tension in the string=T newtons towards right.

So,using newton's 2nd law, T-9.016=4a.............................equation 2    where a is the acceleration of 4 kg block towards right.

Note that magnitude of acceleration for both the blocks is same as they are connected by an inextensible string.

Adding equation 1 and 2,we get,49-9.016=9a=>a=39.984/9 m/s²=4.44 m/s²

Substituting a in equation 1, 49-T=5*39.984/9=>T=26.79 N

if coefficient of friction is 0.3:

Vertical forces on the 5kg block include:

1.gravitational force:mg downwards, where m is mass of block=5 kg and g is gravitational acceleration=9.8 m/s²

So,gravitational force=5*9.8=49 N downwards

2.Force due to tension in the string=T newtons upward.

Using newton's 2nd law:force=mass*acceleration,we get, 49-T=5a........................equation 1 where a is the acceleration of 5kg block,downward.

Since, the 4 kg block is not accelerating vertically, vertical forces must balance=>normal reaction N=weight of the block=mg (where m is mass of the block,g is gravitational acceleration)= 4*9.8=39.2 N

Horizontal forces on the 4 kg block include:

1.friction=kN,where k is coefficient of friction , N is normal reaction

So,friction=0.3*39.2=11.76 N towards left.

2.Tension in the string=T newtons towards right.

So,using newton's 2nd law, T-11.76=4a.............................equation 2 where a is the acceleration of 4 kg block towards right.

Note that magnitude of acceleration for both the blocks is same as they are connected by an inextensible string.

Adding equation 1 and 2,we get,49-11.76=9a=>a=37.24/9 m/s²=4.14 m/s²

Substituting a in equation 1, 49-T=5*37.24/9=>T=28.31 N