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 find the acceleration of the crate, we first analyze the forces acting on it:

1. Identify the Forces:

   • Weight of the crate: $W = mg = 20kg \times 9.81m/s^2 = 196.2N$.
   • Tension in the handle: $T = 40N$.
   • Normal force ($N$) acting vertically upwards.
   • Frictional force ($f$), which is given by $f = \mu N$, where ( \mu = 0.14 ).

2. Resolve Forces:

   • The handle is at an angle $\alpha$. The vertical and horizontal components of tension can be expressed as:
     • $T_x = T \cos(\alpha)$
     • $T_y = T \sin(\alpha)$ where $\sin(\alpha) = \frac{3}{5}$ and $\cos(\alpha) = \frac{4}{5}$, from the triangle.

3. Calculate Normal Force:

   • The normal force can be derived from the balance of vertical forces: [ N + T_y = W \rightarrow N = W - T_y = 196.2N - 40N \times \frac{3}{5} ]
   • This results in: [ N = 196.2N - 24N = 172.2N ]

4. Frictional Force Calculation:

   • The frictional force is: [ f = \mu N = 0.14 \times 172.2N = 24.068N ]

5. Newton's Second Law in the Horizontal Direction:

   • The net force ($F_{net}$) in the horizontal direction is: [ F_{net} = T_x - f ]
   • Plugging in the equations: [ F_{net} = 40N \times \frac{4}{5} - 24.068N = 32N - 24.068N ] [ F_{net} = 7.932N ]

6. Acceleration Calculation:

   • Now applying Newton's second law ($F = ma$): [ a = \frac{F_{net}}{m} = \frac{7.932N}{20kg} = 0.3966 m/s^2 ]
   • Thus, the acceleration of the crate is approximately: [ a \approx 0.3966 m/s^2 ]