A 5 kg block, resting on a rough horizontal table, is connected to a 12 kg block by a light inextensible string that passes over a light frictionless pulley - NSC Physical Sciences - Question 2 - 2017 - Paper 1

In the free-body diagram, the following forces must be included:

• Gravitational force acting downwards: $F_g = m imes g = 5 imes 9.8 = 49 ext{ N}$
• Normal force acting upwards: $F_N$
• Applied force at an angle of 30°: $F_A = 5 ext{ N}$ with components:
  • Horizontal component: $F_{Ax} = F_A imes ext{cos}(30°)$
  • Vertical component: $F_{Ay} = F_A imes ext{sin}(30°)$
• Frictional force opposing the motion: $F_f$

To find the normal force, we can set up the equation for vertical forces:

Since the vertical forces must balance out:

$F_{N} + F_{Ay} - F_g = 0$

Substituting known values:

$F_{N} + 5 imes ext{sin}(30°) - 49 = 0$

Solving for $F_N$:

$F_N = 49 - (5 imes 0.5) = 49 - 2.5 = 46.5 ext{ N}$

The kinetic frictional force can be calculated using:

$F_k = ext{μ}_k imes F_N$

Substituting the coefficient of kinetic friction ($ext{μ}_k = 0.2$) and the normal force:

$F_k = 0.2 imes 46.5 = 9.3 ext{ N}$

Using Newton's Second Law:

$F_{net} = m imes a$

The net force on the block is:

$F_{net} = F_A imes ext{cos}(30°) - F_k$

Here it is:

$F_{net} = 5 imes ext{cos}(30°) - 9.3$

Calculate $F_{net}$:

determining $F_A imes ext{cos}(30°)$: $F_{Ax} = 5 imes 0.866 = 4.33 ext{ N}$

Then, $F_{net} = 4.33 - 9.3 = -4.97 ext{ N}$ Hence, using $F_{net} = m imes a$:

ightarrow a = -0.994 ext{ m/s}^2$$