A learner constructs a push toy using two blocks with masses 1,5 kg and 3 kg respectively - NSC Physical Sciences - Question 2 - 2016 - Paper 1

To calculate the kinetic frictional force ( f_k ) acting on the 3 kg block, we use the formula:

$f_k = \mu_k \times F_N$

Where:

• ( \mu_k = 0.15 ) (coefficient of kinetic friction)
• ( F_N ) is the normal force on the 3 kg block.

In this case, the normal force may be approximated as:

• Assume the 3 kg block is experiencing gravitational force only, thus:
  $F_N = m imes g = 3 kg \times 9.8 m/s^2 = 29.4 N$

Now, substituting into the friction equation:
$f_k = 0.15 \times 29.4 N = 4.41 N$

To draw the free-body diagram for the 1,5 kg block, represent the following forces:

1. The applied force ( F_a ) of 25 N at an angle of 30° to the horizontal.
2. The tension ( T ) in the cord connecting to the 3 kg block.
3. The weight of the block acting downwards: ( W = m \times g = 1.5 kg \times 9.8 m/s^2 = 14.7 N ).
4. The normal force ( F_N ), acting vertically upwards, equal in magnitude to the weight.
5. The kinetic frictional force ( f_k ), acting opposite the direction of motion.

In summary, label each force accordingly in the diagram based on this description.

For the 1.5 kg block, we again use the formula for kinetic friction:

$f_k = \mu_k \times F_N$

Where ( F_N ) is the normal force, which, similar to the previous calculation, equals the weight of the block. So:
$F_N = m \times g = 1.5 kg \times 9.8 m/s^2 = 14.7 N$

Now substituting in the values:
$f_k = 0.15 \times 14.7 N = 2.205 N$

To calculate the tension ( T ) in the cord, we analyze the forces acting on the 1,5 kg block using Newton's Second Law. The equation will take into account the net force the block experiences.

The net force is equal to the applied force minus the frictional force and the tension:

$F_{net} = F_a - f_k - T$

Using the previously calculated values:

• Applied force: ( F_a = 25 N \cos(30°) )
• Frictional force: ( f_k = 2.205 N )
• So we can express it as:

$ma = 25 \cos(30°) - 2.205 - T$

• Let ( a = 1.5 ) (from previous calculations) then we can isolate ( T ) and solve for it, resulting in ( T = 13.19 N ).