Question:

What is the acceleration of block and trolley system shown below, if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? (take g=10 ms^-2).Neglect the mass of the string.

Solution:

Acceleration and tension calculation

The force on the block causes tension in the string and acceleration of the block,

The force of tension acts in the direction opposite to that of the frictional force (in the case of trolley).

Here, the sting is inextensible and the pulley is smooth and the 3 Kg block and the 20 Kg trolley both have the same magnitude of acceleration.

Applying Newton’s second law (F =ma) for the system:

For the block:

The free body diagram of the block and the trolley is shown above. From the free body diagram

$${30-T=3a}$$

This is the equation number (1)

[Force on the block is F=mg =3 Kg X 10 ms-2 = 30 N

F= ma = 3 a]

For the trolley:

Kinetic friction

The kinetic friction depends on nature and conditions of the two surfaces.The range of the kinetic friction is from 0.1 to 1.5)

$${f_k \propto N}$$

$${f_k=\mu_k\ \ N}$$

$${f_k=(0.04) (200 \ N)}$$

$${f_k=8 N}$$

From the free body diagram of the trolley

$${T – f_k = 20a}$$

$${T-8=20a}$$

This is the equation number (2)

Acceleration of the block-trolley system:

Solving the above two equations

$${23a=22 \ \frac{m}{s^2}}$$

$${a=\frac {22}{23}}$$

$${a=0.96 \ \frac{m}{s^2}}$$

Tension in the string:

$${30-T=3a}$$

$${T=30-3a}$$

$${T= 30 -3(0.96)}$$

$${T=30-2.88}$$

$${T=27.1 \ N}$$

Conclusion:

The acceleration of the block-trolley system is 0.96 ms^-2

The tension in the string is 27.1 N