A force F, with a magnitude of 193,19 N, is applied at an angle of 20° with the horizontal to a block of mass 55 kg - NSC Technical Sciences - Question 2 - 2024 - Paper 1

Newton's First Law of Motion states that an object continues in a state of rest or uniform motion (moving with a constant velocity) unless it is acted upon by a net external force. In simpler terms, if no net force acts on an object, its velocity remains constant.

Since the block moves at a constant velocity, the net horizontal force acting on the block must be zero. According to Newton's First Law:

$F_{net} = F - f = 0$ Where F is the applied force and f is the frictional force. Thus, we have: $F = f$ Therefore, the magnitude of the net horizontal force acting on the block is 0 N.

2.4.1 Vertical component of F:

The vertical component of the force F can be calculated using: $F_y = F imes ext{sin}( heta)$ Substituting the values: $F_y = 193.19 imes ext{sin}(20°) = 66.07 \, \text{N}$

2.4.2 Normal force:

The normal force can be calculated as: $N = F_g + F_y$ So, $N = 539.55 \, ext{N} + 66.07 \, ext{N} = 605.62 \, ext{N}$

2.4.3 Frictional force:

The frictional force (f) can be calculated using: $f_k = ext{friction coefficient} (\mu) \times N$ Substituting values: $f_k = 0.3 \times 605.62 = 181.87 \, ext{N}$

When the force F is applied at an angle of 0°, the entire force contributes to moving the block horizontally. In this case, there are no vertical components of F acting upwards, hence the normal force, which is the force balancing the weight of the block, is equal to the weight alone:

$N = F_g = 539.55 \, ext{N}$ Since there is no additional upward force component due to F, the normal force is less than the situation where F is applied at an angle, hence explaining the decrease in the magnitude of the normal force.