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

The kinetic frictional force can be calculated using the formula:

$f_k = \, \mu_k \, \cdot \, N$

where ( N ) is the normal force. The normal force on the 3 kg block, in this case, is equal to its weight because it is on a horizontal surface:

$N = mg = (3 \, \text{kg}) \, \cdot \, (9.8 \, m/s^2) = 29.4 \, N$

Thus, the kinetic frictional force is:

$f_k = 0.15 \, \cdot \, 29.4 \, N = 4.41 \, N$

In the free-body diagram for the 1.5 kg block, the following forces should be included:

1. The applied force (( F_{applied} )), which is 25 N at a 30° angle.
2. The tension in the cord (( T )), pointing to the right.
3. The kinetic frictional force (( f_k )), opposing the direction of motion.
4. The weight of the block (( mg )), acting downwards: $mg = (1.5 \, kg) \, \cdot \, (9.8 \, m/s^2) = 14.7 \, N$
5. The normal force (( N )), acting upwards, equal to the weight. The directions of these forces will help to analyze the motion and solving for acceleration and tension.

To find the kinetic frictional force acting on the 1.5 kg block, we need to calculate the normal force first, which is influenced by the vertical component of the applied force.

The vertical component of the applied force (( F_{applied} )) is:

$F_{y} = F_{applied} \sin(30°) = 25 \, N \cdot 0.5 = 12.5 \, N$

Now, we can calculate the normal force:

$N = mg - F_{y}$ $N = (1.5 \, kg \cdot 9.8 \, m/s^2) - 12.5 \, N = 14.7 \, N - 12.5 \, N = 2.2 \, N$

Then the kinetic frictional force is:

f_k = \, rac{1}{2} \cdot 2.2 \, N = 0.33 \, N

Using Newton's Second Law for the 1.5 kg block:

$F_{net} = T + F_{applied} \cos(30°) - f_k$ Thus, substituting in our known values:

$1.5a = T + 25 \cdot \frac{\sqrt{3}}{2} - 0.33$ Now for the 3 kg block:

$F_{net} = mg - f_k - T$ Where ( f_k ) for the 3 kg block is still calculated as:

$f_k = \, \mu_k \, N = 0.15 \, \cdot \, (3 \cdot 9.8) = 4.41 \, N$ Putting these together:

$T = 13.17 \text{N}$