Two blocks, of weights 3.6 N and 7.2 N, are connectedby a massless string and slide down a 30° inclined plane

Two blocks, of weights 3.6 N and 7.2 N, are connectedby a massless string and slide down a 30° inclined plane.The coefficient of kinetic friction between the lighter block andthe plane is 0.10, and the coefficient between the heavier blockand the plane is 0.20. Assuming that the lighter block leads, find(a) the magnitude of the acceleration of the blocks and (b) thetension in the taut string.

The correct answer and explanation is :

Given:

Weight of the lighter block, ( W_1 = 3.6 \, \text{N} )
Weight of the heavier block, ( W_2 = 7.2 \, \text{N} )
Angle of incline, ( \theta = 30^\circ )
Coefficient of kinetic friction for lighter block, ( \mu_1 = 0.10 )
Coefficient of kinetic friction for heavier block, ( \mu_2 = 0.20 )
Massless string, so tension is the same across both blocks
Both blocks are sliding down the incline, with the lighter block leading.

Assumptions:

The system is in motion, so kinetic friction acts on both blocks.
The gravitational force component pulling the blocks down the incline is ( W \sin \theta ).
The frictional force is given by ( f = \mu N ), where ( N ) is the normal force acting on the blocks.

Step 1: Find the masses of the blocks

Since weight ( W = mg ), the mass of each block can be found by dividing the weight by ( g = 9.8 \, \text{m/s}^2 ).

For the lighter block:
[
m_1 = \frac{W_1}{g} = \frac{3.6}{9.8} \approx 0.367 \, \text{kg}
]

For the heavier block:
[
m_2 = \frac{W_2}{g} = \frac{7.2}{9.8} \approx 0.735 \, \text{kg}
]

Step 2: Forces on each block

Lighter block (Block 1):

Gravitational force down the incline: ( W_1 \sin \theta = 3.6 \sin 30^\circ = 3.6 \times 0.5 = 1.8 \, \text{N} )
Normal force: ( N_1 = W_1 \cos \theta = 3.6 \cos 30^\circ = 3.6 \times 0.866 \approx 3.12 \, \text{N} )
Frictional force: ( f_1 = \mu_1 N_1 = 0.10 \times 3.12 = 0.312 \, \text{N} )

Thus, the net force on Block 1 is:
[
F_{\text{net1}} = W_1 \sin \theta – f_1 = 1.8 – 0.312 = 1.488 \, \text{N}
]

Heavier block (Block 2):

Gravitational force down the incline: ( W_2 \sin \theta = 7.2 \sin 30^\circ = 7.2 \times 0.5 = 3.6 \, \text{N} )
Normal force: ( N_2 = W_2 \cos \theta = 7.2 \cos 30^\circ = 7.2 \times 0.866 \approx 6.24 \, \text{N} )
Frictional force: ( f_2 = \mu_2 N_2 = 0.20 \times 6.24 = 1.248 \, \text{N} )

Thus, the net force on Block 2 is:
[
F_{\text{net2}} = W_2 \sin \theta – f_2 = 3.6 – 1.248 = 2.352 \, \text{N}
]

Step 3: Apply Newton’s Second Law

Let ( T ) be the tension in the string and ( a ) be the acceleration of the system.

For the lighter block (Block 1):
[
F_{\text{net1}} = m_1 a = 1.488
]
[
a = \frac{1.488}{0.367} \approx 4.06 \, \text{m/s}^2
]

For the heavier block (Block 2):
[
F_{\text{net2}} = m_2 a = 2.352
]
[
a = \frac{2.352}{0.735} \approx 3.20 \, \text{m/s}^2
]

Since both blocks are connected by a string, they will accelerate at the same rate. Therefore, we need to solve the system for a common acceleration.

Step 4: Solve for the common acceleration

Using the net forces, we have the following equations:

( T = m_1 a + f_1 )
( T = m_2 a + f_2 )

Equating the two:
[
m_1 a + f_1 = m_2 a + f_2
]
[
0.367 a + 0.312 = 0.735 a + 1.248
]
Solving for ( a ):
[
0.367 a – 0.735 a = 1.248 – 0.312
]
[
-0.368 a = 0.936
]
[
a = – \frac{0.936}{0.368} \approx 2.55 \, \text{m/s}^2
]

Step 5: Find the tension in the string

Now that we have the acceleration, we can calculate the tension ( T ) in the string. Using the equation for the lighter block:
[
T = m_1 a + f_1 = 0.367 \times 2.55 + 0.312 = 0.935 + 0.312 = 1.247 \, \text{N}
]

Final Answers:

(a) The magnitude of the acceleration is approximately ( 2.55 \, \text{m/s}^2 ).

(b) The tension in the string is approximately ( 1.25 \, \text{N} ).