A2 June 2025 Q4

EdexcelCurrent spec7 marksWork-Energy Principle

Figure 1: a ramp inclined at angle θ to the horizontal, with A lower and B higher on the ramp, AB = 18 m, and the package projected up the ramp from A at 8 m s⁻¹ — Figure 1

A rough straight ramp is fixed to horizontal ground. The ramp is inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{1}{14}\). The points \(A\) and \(B\) are on a line of greatest slope of the ramp with \(AB = 18\) m and \(B\) above \(A\), as shown in Figure 1.

A package of mass 0.5 kg is projected up the ramp from \(A\) with speed \(8\ \text{m s}^{-1}\) and comes to instantaneous rest at \(B\).

The work done against friction as the package moves from \(A\) to \(B\) is \(W\) joules.

The package is modelled as a particle and air resistance is ignored.

(a) Use the work-energy principle to show that \(W = 9.7\) (3)

The coefficient of friction between the package and the ramp is \(\mu\)

(b) Find the value of \(\mu\) (4)

Mark scheme (a)

Scheme	Marks	AO
Work-energy equation:	M1	3.4
\(\dfrac{1}{2} \times 0.5 \times 8^2 = W + 0.5g \times 18\sin\theta\)	A1	1.1b
\(W = 16 - 9g \times \dfrac{1}{14} = 9.7\) *	A1*	2.2a
	(3)

Notes

M1: Form work-energy equation in terms of \(W\) (and \(g\) and \(\theta\)) only. All required terms present and no extras. All terms dimensionally correct (of correct structure). Condone \(\pm\) sign errors on terms and sin/cos confusion on vertical height. M0 if a term is missing or for incorrect trig use eg \(18\tan\theta,\ \dfrac{18}{\sin\theta},\ \dfrac{18}{\cos\theta}\). M0 for use of suvat

A1: Correct unsimplified equation, no need to replace trig.

A1*: Obtain given answer from complete and correct working. Must see a line of working between the initial equation and the given answer. Condone missing \(W\) during working but must see ‘\(W = \ldots\)’ for the final mark.

Mark scheme (b)

Scheme	Marks	AO
Use of \(F = \mu R = \mu \times 0.5g\cos\theta\)	M1	3.1b
Complete method to form a dimensionally correct equation in \(\mu\) (and \(\theta\)) using Either work done \(= F \times 18\) Or suvat and N2L	M1	3.4
Either \(9.7 = \mu \times 0.5g\cos\theta \times 18\) Or relevant suvat to find \(a\) \(\left(= -\dfrac{16}{9}\right)\) and N2L to form \(-\mu \times 0.5g\cos\theta - 0.5g\sin\theta = 0.5\left(-\dfrac{16}{9}\right)\)	A1	1.1b
\(\mu = 0.11 \quad (0.110)\)	A1	1.1b
	(4)
	(7 marks)

Notes

M1: Correct use of \(F = \mu R\) and \(R = 0.5g\cos\theta\) to form an expression for Friction. Dimensionally correct. Condone sin/cos confusion. Missing \(g\) is an accuracy error not a method error.

M1: Complete method to form a dimensionally correct equation in \(\mu\) (\(\theta\) and \(g\)). Trig does not need to be replaced for M mark. M0 for \(W = \mu R\).
May use 9.7 and work done \(= F \times 18\) to form a dimensionally correct equation in \(\mu\).
May see relevant suvat to find acceleration, followed by N2L to form dimensionally correct equation \(\mu\). N2L must contain all relevant terms and no extras. Condone sin/cos confusion on the weight components.

A1: Correct unsimplified equation in \(\mu\) (and \(g\)) with trig replaced correctly.

A1: 2 sf or 3 sf only. A0 for use of \(g = 9.81\ \text{m s}^{-2}\)

(Corrected from the printed mark scheme: the N2L equation in the second method is printed as \(-\mu \times 0.5g\cos\theta \times 18 - 0.5g\sin\theta = 0.5\left(-\dfrac{16}{9}\right)\); the friction force has no factor of 18 in an equation of motion.)