Mathematics 9709/41 — May/June 2015

Cambridge AS Level · Mechanics · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Forces and Equilibrium · Energy, Work and Power · Newton's Laws of Motion · Kinematics of Motion in a Straight Line

Q1Forces and EquilibriumEnergy, Work and PowerFree sample

A block $B$ of mass $2.7\text{ kg}$ is pulled at constant speed along a straight line on a rough horizontal floor. The pulling force has magnitude $25\text{ N}$ and acts at an angle of $\theta$ above the horizontal. The normal component of the contact force acting on $B$ has magnitude $20\text{ N}$ .

(i)

Show that $\sin \theta = 0.28$ .

DifficultyMedium-Easy

Worked solution

Approach

The block moves horizontally at constant speed, so its vertical acceleration is zero. Resolve the forces vertically, taking upward as positive, and set the upward forces equal to the block's weight. The normal component of the contact force is the upward normal reaction $R = 20\text{ N}$ .

Working

Taking $g = 10\text{ m s}^{-2}$ , the weight of $B$ is

W = mg = 2.7 \times 10 = 27\text{ N}.

The vertical component of the $25\text{ N}$ pulling force is $25\sin\theta$ upward. Resolving vertically:

R + 25\sin\theta = mg

20 + 25\sin\theta = 27

25\sin\theta = 7

\sin\theta = \frac{7}{25} = 0.28.

Therefore $\sin\theta = 0.28$ , as required.

Answer

\sin\theta = 0.28

Final answer

sin θ = 0.28

Detailed explanation

Walkthrough

The block is pulled along a horizontal rough floor at constant speed. Constant speed in a straight line means the block has no acceleration, and in particular no vertical acceleration; so the vertical forces must balance. The forces with vertical components are the weight $mg = 2.7g$ downward, the normal component of the contact force $R = 20\text{ N}$ upward, and the vertical component of the pulling force $25\sin\theta$ upward. Resolving vertically gives $R + 25\sin\theta = mg$ . Since the question uses the standard value $g = 10\text{ m s}^{-2}$ , the weight is $27\text{ N}$ . Substitute $R = 20$ and solve: $25\sin\theta = 7$ , so $\sin\theta = 7/25 = 0.28$ . This is a 'show that' question, so the final line must be displayed clearly.

Key Takeaways

This part tests resolving forces into components and applying the equilibrium condition in a direction in which there is no acceleration. It also shows that the normal reaction is not necessarily equal to the weight: the upward component of the pulling force reduces the normal contact force. The friction force is horizontal and does not enter the vertical resolution.

Common Mistakes

A common error is to write $R = mg$ immediately, forgetting that the vertical component of the pull also supports part of the weight. Another is to use the horizontal component $25\cos\theta$ in the vertical equation. Students also sometimes forget to state $g = 10$ ; with a different value of $g$ the result would not be $0.28$ .

Things to Be Careful About

The angle $\theta$ is measured above the horizontal, so the vertical component is $25\sin\theta$ and the horizontal component is $25\cos\theta$ . The block is on a rough floor, but friction acts horizontally and does not affect the vertical resolution. Since the block moves at constant speed along a straight line, its vertical acceleration is zero, so vertical forces are in equilibrium.

Techniques used

resolve forces verticallyapply the vertical equilibrium conditionsolve for the sine of an angle

(ii)

Find the work done by the pulling force in moving the block a distance of $5\text{ m}$ .

DifficultyMedium-Easy

Worked solution

Approach

The block moves horizontally, so only the horizontal component of the pulling force does work. Use the work formula $W = Fd\cos\theta$ , and find $\cos\theta$ from $\sin\theta = 0.28$ using the identity $\cos^2\theta + \sin^2\theta = 1$ .

Working

\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - 0.28^2}

\cos\theta = \sqrt{1 - 0.0784} = \sqrt{0.9216} = 0.96.

The work done by the pulling force over a horizontal distance of $5\text{ m}$ is

W = Fd\cos\theta = 25 \times 5 \times 0.96

W = 125 \times 0.96 = 120\text{ J}.

The vertical component of the pulling force does no work because there is no vertical displacement.

Answer

120\text{ J}

Final answer

120 J

Detailed explanation

Walkthrough

The work done by a constant force is the product of the component of the force in the direction of motion and the distance moved. The block moves horizontally, so the component of the pulling force along the motion is $25\cos\theta$ ; the vertical component $25\sin\theta$ does no work because there is no vertical displacement. From part (i), $\sin\theta = 0.28$ . Since $\theta$ is acute, $\cos\theta$ is positive, and $\cos\theta = \sqrt{1 - 0.28^2} = 0.96$ . Therefore the work done is $W = 25 \times 5 \times 0.96 = 125 \times 0.96 = 120\text{ J}$ . The answer is positive because the horizontal component of the force is in the direction of motion.

Key Takeaways

This part applies the work formula $W = Fd\cos\theta$ to a force at an angle to the displacement. It also reinforces the link between the trigonometric components of a force and the Pythagorean identity. Work is a scalar measured in joules; a force component perpendicular to the displacement does no work.

Common Mistakes

A frequent mistake is to use $\sin\theta$ instead of $\cos\theta$ in the work formula. Another is to compute the work as $25 \times 5 = 125\text{ J}$ , which ignores that the force is not parallel to the displacement. Some students also forget to find $\cos\theta$ from $\sin\theta$ and instead leave an expression involving $\theta$ .

Things to Be Careful About

Use the distance moved by the block, $5\text{ m}$ , and the full force magnitude, $25\text{ N}$ , with the cosine of the angle between the force and the displacement. The vertical component of the pulling force does no work because the displacement is horizontal. Since $\theta$ is above the horizontal, $\cos\theta = 0.96$ is positive. The final work is $120\text{ J}$ , not $125\text{ J}$ .

Techniques used

resolve force into component parallel to displacementuse the Pythagorean identity to find cosineapply the work formula W = Fd cos θ

The rest of this paper

6 more questions

Q2Forces and Equilibrium5M
Q3Forces and Equilibrium · Newton's Laws of Motion5M
Q4Energy, Work and Power6M
Q5Newton's Laws of Motion · Energy, Work and Power8M
Q6Kinematics of Motion in a Straight Line10M
Q7Newton's Laws of Motion · Forces and Equilibrium · Energy, Work and Power12M

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