Photo AI
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
Question 7
A wooden crate of mass 20kg is pulled in a straight line along a rough horizontal floor using a handle attached to the crate. The handle is inclined at an angle \( ... show full transcript
Worked Solution & Example Answer: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
Step 1
find the acceleration of the crate.
Answer
To find the acceleration of the crate, we will start by analyzing the forces acting on it. Given:
- Mass of the crate, ( m = 20 \text{ kg} )
- Force applied using the handle, ( F = 40 \text{ N} )
- Coefficient of friction, ( , ext{friction} = 0.14 )
Step 1: Calculate the angle ( \alpha )
From the tangent value given:
[ \tan \alpha = \frac{3}{4} ]
Using this, we can find the sine and cosine values:
[ \sin \alpha = \frac{3}{5}, \quad \cos \alpha = \frac{4}{5} ]
Step 2: Determine the normal force ( N )
The normal force can be expressed as:
[ N = mg - T\sin \alpha ]
Where ( T ) is the tension in the handle. So:
[ N = 20g - 40 \times \frac{3}{5} = 20g - 24
N = 20g - 24 = 20(9.81) - 24 \approx 196.2 - 24 = 172.2 \text{ N} ]
Step 3: Calculate the frictional force ( F_f )
[ F_f = ext{friction} \cdot N = 0.14 \times 172.2 \approx 24.068 \text{ N} ]
Step 4: Apply Newton's second law
The net force ( F_{net} ) acting on the crate: [ F_{net} = F - F_f = 40 - 24.068 = 15.932 \text{ N} ] Using Newton's second law, we find the acceleration ( a ): [ a = \frac{F_{net}}{m} = \frac{15.932}{20} \approx 0.7966 \text{ m/s}^2 ]
Thus, the acceleration of the crate is approximately ( 0.80 \text{ m/s}^2 ).
Step 2
Explain briefly why the acceleration of the crate would now be less than the acceleration of the crate found in part (a).
Answer
The acceleration of the crate in part (b) would now be less than in part (a) due to two main reasons:
-
Increased Frictional Effects: When the crate is pushed along the rough surface, the normal force exerted on the crate changes. In part (a), the tension in the handle provided an upward force which counteracted some of the weight of the crate, effectively lowering the normal force and thus the frictional force. In part (b), while the angle remains the same, the distribution of forces means that the frictional force remains higher relative to the total applied force due to changes in angle and direction of applied force.
-
Constant Thrust Force: Although the thrust remains the same (40N), if we consider the force balance, we know that with increased friction and less effective distribution in the pushing direction, less of the thrust can be used to accelerate the crate, leading to a lower acceleration compared to that in part (a).
In summary, friction becomes a larger opposing force in part (b), reducing the net force acting on the crate and therefore its acceleration.