A wooden crate of mass 20kg is pulled in a straight line along a rough horizontal floor using a handle attached to the crate - Edexcel - A-Level Maths Mechanics - Question 7 - 2018 - Paper 2

To calculate the acceleration, we first analyze the forces acting on the crate.

1. Break down the forces: The vertical component of the tension is given by: $T_y = T \sin \alpha$ and the horizontal component is: $T_x = T \cos \alpha$

2. Frictional force: The frictional force ( F_f ) can be calculated using: $F_f = ext{friction coefficient} \times N$ where ( N ) is the normal force given by: $N = mg - T_y$

   Substituting the values:

   • Mass ( m = 20 , ext{kg} )
   • Gravitational acceleration ( g = 9.81 , ext{m/s}^2 )
   • ( T = 40 , ext{N} )
   • Use ( \tan \alpha = \frac{3}{4} ) implies ( \sin \alpha = \frac{3}{5} ) and ( \cos \alpha = \frac{4}{5} )

   Hence, the normal force: $N = 20 \times 9.81 - 40 \times \frac{3}{5} = 196.2 - 24 = 172.2 \, \text{N}$ Thus, frictional force becomes: $F_f = 0.14 \times 172.2 = 24.108 \, \text{N}$

3. Net horizontal force: The net force acting horizontally on the crate is:

   = 32 - 24.108 = 7.892 \, \text{N}$$

4. Acceleration: Using Newton's second law: $F_{net} = ma$ thus: $a = \frac{F_{net}}{m} = \frac{7.892}{20} = 0.3946 \, \text{m/s}^2$

Thus, the acceleration of the crate is approximately ( 0.3946 , \text{m/s}^2 ).