Two blocks, A and B, of masses 25 kg and 45 kg respectively, joined by a light inextensible string, are moved towards the east on a rough horizontal surface - NSC Technical Sciences - Question 3 - 2021 - Paper 1

The free-body diagram for block A includes the following forces:

• Applied force, $F_A = 50 N$, acting horizontally to the right.
• Gravitational force, $F_g = mg = 25 kg \times 9.8 m/s^2 = 245 N$, acting vertically downward.
• Normal force, $N$, acting vertically upward.
• Kinetic frictional force, $F_{fA} = 5.82 N$, acting horizontally to the left.

Label the arrows representing these forces accordingly.

Newton's Second Law of Motion states that when a net force acts on an object, it accelerates in the direction of the net force. The acceleration is directly proportional to the net force and inversely proportional to the mass of the object.

The coefficient of kinetic friction, $\, \mu_k$, can be calculated using:

$f_k = \mu_k \times N$

Here, the normal force $N$ for block A is equal to the gravitational force:

$N = F_g = 245 N$

The net force acting on block A is:

$F_{net} = F_A - f_k = 50 N - 5.82 N$

Now, substituting into the equation:

$\mu_k = \frac{F_{net}}{N} = \frac{50 N - 5.82 N}{245 N} = \frac{44.18 N}{245 N} \approx 0.180$

Let east be positive. For block A:

$F_{netA} = ma$ $T + F_A \cos(28^{\circ}) - f_{kA} = m_A a$

For block B:

$F_{netB} = ma$ $T - F_{B} + f_{kB} = m_B a$

Set the two equations equal and solve for $T$:

For block A, substituting values:

$T + 50 N \times \cos(28^{\circ}) - 5.82 N = 25a$

For block B:

$T - 350 N + 8.35 N = 45a$

By solving these simultaneous equations, we can find the tension in the string, $T$.