In this question use $g = 9.8 \, ext{ms}^{-2}$
A rough wooden ramp is 10 metres long and is inclined at an angle of 25° above the horizontal - AQA - A-Level Maths Pure - Question 19 - 2022 - Paper 2
Question 19
In this question use $g = 9.8 \, ext{ms}^{-2}$
A rough wooden ramp is 10 metres long and is inclined at an angle of 25° above the horizontal. The bottom of the ram... show full transcript
Worked Solution & Example Answer:In this question use $g = 9.8 \, ext{ms}^{-2}$
A rough wooden ramp is 10 metres long and is inclined at an angle of 25° above the horizontal - AQA - A-Level Maths Pure - Question 19 - 2022 - Paper 2
Step 1
19 (a) The tension in the rope is 230 N. The crate accelerates up the ramp at 1.2 m/s². Find the coefficient of friction between the crate and the ramp.
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Answer
To find the coefficient of friction, we first need to resolve the forces acting on the crate along the ramp and perpendicular to the ramp.
Resolve Weight: The weight of the crate can be resolved into two components:
Parallel to the ramp: Wextparallel=mgextsinheta
Perpendicular to the ramp:
Wextperpendicular=mgextcosheta
Here, ( m = 20 , ext{kg} ), ( g = 9.8 , ext{ms}^{-2} ), and ( heta = 25° ).
Therefore,
Wextparallel=20imes9.8imesextsin(25°)Wextperpendicular=20imes9.8imesextcos(25°)
Applying Newton's Second Law:
The net force equation along the ramp can be written as:
T−Wextparallel−Fextfriction=ma
Where:
Thus, substituting into the equation:
230−(20imes9.8imesextsin(25°))−extμ(20imes9.8×extcos(25°))=20(1.2)
Solve for extμ:
After substitution and simplification, we can find ( ext{μ} )
Rearranging to solve for ( ext{μ} ):
230−196extsin(25°)−20extμ(9.8extcos(25°))=24
This leads to an expression to calculate ( ext{μ} ) yielding ( ext{μ} , ext{≈} , 0.69 ).
Step 2
19 (b) (i) The crate takes 3.8 seconds to reach the top of the ramp. Find the distance OA.
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Answer
Using the kinematic equation for uniformly accelerated motion,
s=ut+21at2
where:
( u = 0 ) (initial velocity)
( a = 1.2 , ext{ms}^{-2} )
( t = 3.8 , ext{s} )
Substituting these values into the equation:
s=0(3.8)+21(1.2)(3.82)
= \frac{1}{2}(1.2)(14.44) \approx 8.664 , ext{m} $$
Therefore, the distance OA is approximately ( 10 , ext{m} - 8.664 , ext{m} = 1.336 , ext{m} ).
Step 3
19 (b) (ii) Other than air resistance, state one assumption you have made about the crate in answering part (b)(i).
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Answer
One assumption made is that the crate is treated as a particle, meaning we neglect its dimensions and consider all its mass concentrated at a point. This simplifies the dynamics involved in the motion up the ramp.