AS Forces and MotionEnergy & PowerMaterials 3.1.1(a)3.1.1(b)3.3.1(a)3.3.1(b)3.3.1(c)

~15 min Difficulty: 5/10 10 marks

Prior knowledge Basic SUVATdefinition of weightenergy forms.

Problem structure

(a) 3 marks, Numerical, Kinematics (3.1.1a). (b) 5 marks, Numerical, Conservation of Energy and Hooke's Law (3.3.1c, 3.4.1b/c). (c) 2 marks, Numerical/Qualitative, Work Done (3.3.1a/b).

Solve the problem

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A testing rig uses a heavy block of mass $M = 12.0\text{ kg}$ to calibrate a vertical safety spring. The block is initially held at rest against the top of the spring. When released, the block compresses the spring. (a) The block is released from rest and undergoes a constant acceleration of $a = 8.20\text{ m s}^{-2}$ downwards for the first $0.150\text{ s}$ before it makes full contact with the spring's main resistance. Calculate the displacement of the block during this initial time interval and its instantaneous velocity at $t = 0.150\text{ s}$. [3 marks] (b) After the initial $0.150\text{ s}$, the block begins to compress the spring. The spring obeys Hooke’s Law and has a force constant $k = 4500\text{ N m}^{-1}$. The block comes to a momentary rest when the spring has been compressed by a distance $x$. By considering the principle of conservation of energy from the moment the block was first released, calculate the maximum compression $x$ of the spring. Assume air resistance is negligible and the gravitational potential energy is lost over the total vertical distance moved. [5 marks] (c) A sensor records the force $F$ exerted by the spring on the block. Calculate the work done by the spring on the block during the compression $x$ found in part (b). State the direction of this work relative to the displacement of the block. [2 marks]

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Your answer

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A testing rig uses a heavy block of mass $M = 12.0\text{ kg}$ to calibrate a vertical safety spring. The block is initially held at rest against the top of the spring. When released, the block compresses the spring.

(a) The block is released from rest and undergoes a constant acceleration of $a = 8.20\text{ m s}^{-2}$ downwards for the first $0.150\text{ s}$ before it makes full contact with the spring's main resistance. Calculate the displacement of the block during this initial time interval and its instantaneous velocity at $t = 0.150\text{ s}$.

[3]

(b) After the initial $0.150\text{ s}$, the block begins to compress the spring. The spring obeys Hooke’s Law and has a force constant $k = 4500\text{ N m}^{-1}$. The block comes to a momentary rest when the spring has been compressed by a distance $x$. By considering the principle of conservation of energy from the moment the block was first released, calculate the maximum compression $x$ of the spring. Assume air resistance is negligible and the gravitational potential energy is lost over the total vertical distance moved.

[5]

(c) A sensor records the force $F$ exerted by the spring on the block. Calculate the work done by the spring on the block during the compression $x$ found in part (b). State the direction of this work relative to the displacement of the block.

[2]

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(a) Using equations for constant acceleration: Displacement $s = ut + \frac{1}{2}at^2$. Since $u = 0$: $s = 0 + 0.5 \times 8.20 \times (0.150)^2$ $s = 0.09225\text{ m} = 9.23 \times 10^{-2}\text{ m}$ (or $9.23\text{ cm}$). Instantaneous velocity $v = u + at$: $v = 0 + 8.20 \times 0.150$ $v = 1.23\text{ m s}^{-1}$. (b) Total vertical distance fallen $h = s + x$, where $s = 0.09225\text{ m}$. By Conservation of Energy: Loss in Gravitational Potential Energy (GPE) = Gain in Elastic Potential Energy (EPE). $Mg(s + x) = \frac{1}{2}kx^2$ Substitute known values ($g = 9.81\text{ m s}^{-2}$): $12.0 \times 9.81 \times (0.09225 + x) = 0.5 \times 4500 \times x^2$ $117.72(0.09225 + x) = 2250x^2$ $10.86 + 117.72x = 2250x^2$ Rearrange into a quadratic equation $ax^2 + bx + c = 0$: $2250x^2 - 117.72x - 10.86 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{117.72 \pm \sqrt{(-117.72)^2 - 4(2250)(-10.86)}}{2(2250)}$ $x = \frac{117.72 \pm \sqrt{13858 + 97740}}{4500}$ $x = \frac{117.72 \pm 334.06}{4500}$ Since $x$ must be positive: $x = \frac{451.78}{4500} = 0.10039\dots\text{ m}$ $x \approx 0.100\text{ m}$ (to 3 SF). (c) Work done by the spring $W = \text{Area under } F\text{-}x \text{ graph} = \frac{1}{2}kx^2$. $W = 0.5 \times 4500 \times (0.10039)^2 = 22.67\dots\text{ J}$ $W \approx 22.7\text{ J}$. Direction: The force of the spring is upwards (opposing compression) while the displacement is downwards. Therefore, the work done by the spring on the block is negative (or "opposes the motion"). Final Answers: (a) $s = 0.0923\text{ m}$, $v = 1.23\text{ m s}^{-1}$ (b) $x = 0.100\text{ m}$ (c) $W = 22.7\text{ J}$, direction is opposite to displacement.

Marking guidance

(a) M1: Correct use of $s = \frac{1}{2}at^2$ to find $0.09225\text{ m}$. M1: Correct use of $v = at$ to find $1.23\text{ m s}^{-1}$. A1: Both answers given to 3 significant figures with correct units. (b) M1: Identification of total height $h = s + x$ (must include the initial drop). M1: Equating GPE loss to EPE gain: $Mg(s+x) = \frac{1}{2}kx^2$. M1: Correct substitution of values into a quadratic form. M1: Correct application of the quadratic formula or solver. A1: Final answer $x = 0.100\text{ m}$ (accept $0.10$ or $0.100$). (Note: If $s$ was omitted from GPE, $x = \sqrt{2Mgx/k}$ is incorrect physics; if $s$ was used as the only height, max 2 marks for EPE calculation). (c) M1: Calculation of $\frac{1}{2}kx^2$ using their $x$ from (b). A1: Correct value ($\approx 22.7\text{ J}$) and stating the work is negative or in the opposite direction to displacement.