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 can start by resolving the forces acting on it.

1. Identify Forces: The tension in the handle ($T = 40N$) acts at an angle $\alpha$ with the horizontal. The horizontal component of the tension is given by:

   $T_x = T \cos(\alpha)$

   The vertical component of the tension is:

   $T_y = T \sin(\alpha)$

2. Calculate Components: Given that $\tan(\alpha) = \frac{3}{4}$, we can use:

   $\sin(\alpha) = \frac{3}{5}, \quad \cos(\alpha) = \frac{4}{5}$

   Thus,

   $T_x = 40 \cdot \frac{4}{5} = 32N, \quad T_y = 40 \cdot \frac{3}{5} = 24N$

3. Weight and Normal Force: The weight of the crate $W = mg = 20kg \cdot 9.81m/s^2 = 196.2N$. The normal force $N$ is then:

   $N = W - T_y = 196.2N - 24N = 172.2N$

4. Frictional Force: The frictional force $F_r$ can be calculated using the coefficient of friction ($\mu = 0.14$):

   $F_r = \mu N = 0.14 \cdot 172.2N \approx 24.1N$

5. Net Force: The net force $F_{net}$ acting on the crate in the horizontal direction is:

   $F_{net} = T_x - F_r = 32N - 24.1N = 7.9N$

6. Acceleration: Finally, using Newton's second law, we can find the acceleration $a$:

   $a = \frac{F_{net}}{m} = \frac{7.9N}{20kg} = 0.395 m/s^2$

Thus, the acceleration of the crate is approximately 0.40 m/s².