Answer sheet

Practice: Work-energy theorem

1. A person drives 8.0 km [N], and then 6.0 km [W]. Find the total displacement.

Suggested answer:

Solution: Using the Pythagorean theorem.

Given

$\vec{d_{1}} = 8.0 km [N]$

$\vec{d_{2}} = 6.0 km [W]$

Required

$\vec{d_{T}} = ?$

Analysis and Solution

${|\vec{Δ d}|}^{2} = {(8.0 km)}^{2} + {(6.0 km)}^{2}$

$|\vec{Δ d}| = \sqrt{{(8.0 km)}^{2} + {(6.0 km)}^{2}}$

$|\vec{Δ d}| = 1.0 \times 10^{1} km$

$\tan θ = \frac{6.0 km}{8.0 km}$

$\tan θ = 0.75$

$θ = \tan^{- 1} (0.75)$

$θ = {36.9}^{°}$

Statement

∴ The resultant displacement is 1.0 × 10¹ km [N 37° W].

2. A person travels 2.0 m [E 20° S], then 4.0 m [S]. Find the total displacement.

Suggested answer:

Solution: Using sine and cosine laws.

${|\vec{d_{T}}|}^{2} = {(2.0 m)}^{2} + {(4.0 m)}^{2} - 2 (2.0 m) (4.0 m) \cos 1 10^{\circ}$

${|\vec{d_{T}}|}^{2} = 25.47 m^{2}$

$|\vec{d_{T}}| = 5.0 m$

$\frac{\sin θ}{4.0 m} = \frac{\sin 1 10^{\circ}}{5.0 m}$

$\sin θ = 0.752$

$θ = 48^{\circ}$

Based on the diagram above, the 20° is added to get 68°.

The resultant displacement is 5.0 m [E68°S] or [S22°E].

3. A truck drives 1.0 × 10² km [S], turns and drives 80.0 km [W 30° S], then turns again and drives 20.0 km [N]. Find the total displacement.

Suggested answer:

Solution: Using perpendicular components method.
[North] and [East] directions are positive.

$\vec{d_{1 y}} = - 100 km$

$\vec{d_{2 y}} = - (80 km) (\sin 3 0^{\circ}) = - 40 km$

$\vec{d_{3 y}} = 20 km$

$\vec{d_{y}} = - 100 km + (- 40 km) + 20 km = - 120 km$

$\vec{d_{2 x}} = (-80 km) (\cos 3 0^{\circ}) = 69.3 km$

$|\vec{d_{T}}| = \sqrt{{|\vec{d_{x}}|}^{2} + {|\vec{d_{y}}|}^{2}}$

$|\vec{d_{T}}| = \sqrt{{(69.3 km)}^{2} + {(- 120 km)}^{2}}$

$|\vec{d_{T}}| = 139 km$

$\tan θ = \frac{120 km}{69.3 km}$

$\tan θ = 1.73$

$θ = 60^{\circ}$

The resultant displacement is 140 km [S30°W].

4. Resolve the following vectors into their components.

17 m/s [N]
40 m/s [S 45° E]

Suggested answer:

a. 17 m/s [N]

Given

$\vec{v}$ = 17 m/s [N]

Required

$\vec{v_{x}}$ = ?

$\vec{v_{y}}$ = ?

Solve

Draw a vector diagram that includes a directional label to represent the x and y components as shown in the following diagram.

Since vector falls along the north-south vertical line, there is no horizontal component. The vector only has a vertical component.

The vector only has a vertical- component, $\vec{v_{y}}$ = 17 m/s [N]. There is no horizontal- component, $\vec{v_{x}}$ of this vector is 0 m/s [E].

b. 40 m/s [S45°E]

Given

$\vec{v}$ = 40 m/s [S 45°E]

Required

$\vec{v_{x}}$ = ?

$\vec{v_{y}}$ = ?

Solve

Draw a vector diagram that includes a directional label to represent the x and y components as shown in the following diagram.

Use the correct trigonometric formula to resolve each component.

For the horizontal-component use the sin formula:

$\sin θ = \frac{v_{x}}{v}$

Rearranging and substituting:

$v_{x} = v \sin θ$

$v_{x} = (40 m / s) (\sin 4 5^{\circ})$

$v_{x} = 28.28 m / s$ (2 extra digits)

$v_{x} = 28 m / s [E]$

For the vertical-component use the cosine formula:

$\cos θ = \frac{v_{y}}{v}$

Rearranging and substituting:

$v_{y} = v \cos θ$

$v_{y} = (40 m / s) {(\cos 4 5)}^{\circ}$

$v_{y} = 28.28 m / s$ (2 extra digits)

$v_{y} = 28 m / s [S]$

Therefore, the horizontal-component, $\vec{v_{x}}$ of this vector is 30 m/s [E] and the vertical-component, $\vec{v_{y}}$ of this vector is 30 m/s [S].

Note that "40" has just one significant digit, so our components should as well.

5. Resolve this vector into its components.
25 m/s [E]

Suggested answer:

6. Resolve this vector into its components.
95 m/s [N 20° W]

Given:

$\vec{v}$ = 95 m/s [N 20°W]

Required:

$\vec{v_{x}}$ = ?

$\vec{v_{y}}$ = ?

Analysis:

Draw a vector diagram that includes a directional label to represent the x and y components.

Use the correct trigonometric formula to resolve each component.

Solution:

The following diagram is representing the scenario.

For the horizontal-component use the sine formula:

$\sin θ = \frac{v_{x}}{v}$

Rearranging and substituting:

$v_{x} = v \sin θ$

$v_{x} = (95 m / s) (\sin 2 0^{\circ})$

$v_{x} = 32.49 m / s$ (2 extra digits)

$v_{x} = 32 m / s [W]$

For the vertical-component use the cosine formula:

$\cos θ = \frac{v_{y}}{v}$

Rearranging and substituting:

$v_{y} = v \cos θ$

$v_{y} = (95 m / s) {(\cos 2 0)}^{\circ}$

$v_{y} = 89.27 m / s$ (2 extra digits)

$v_{y} = 89 m / s [N]$

Summary:

Therefore, the horizontal-component, $\vec{v_{x}}$ of this vector is 32 m/s [W] and the vertical-component, $\vec{v_{y}}$ of this vector is 89 m/s [N].

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