Category Archives: Engineering Mathematics Blog

Part A: To convert the given mass in kilogram to newton force, we simply need to multiply by the acceleration due to gravity of 9.81 m/s². We need to take into account that $1\:\text{kg m/s}^2\:=1\:\text{N}$ .

\begin {aligned} 8\:\text{kg} & =8\:\text{kg}\times 9.81\:\text{m/s}^2 \\ &=78.48\:\text{N} \end {aligned}

Part B: Using the same principle from Part A, we have

\begin {aligned} 0.04\:\text{kg}&=0.04\:\text{kg}\times 9.81\:\text{m/s}^2\\ &=0.3924\:\text{N} \end {aligned}

Part C: So, we are given 760 Mg (megagram). We know that 1 Mg is equivalent to 1000 kg. Therefore, 760 Mg is equal to 760,000 kg. Therefore, we have

\begin {aligned} 760\:000\:\text{kg}&=760\:000\:\text{kg}\times 9.81\:\text{m/s}^2\\ &=7\:455\:600\:\text{N} \end{aligned}

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Physics

College Physics by Openstax

Physics for Scientists and Engineers: A Strategic Approach Third Edition by Randall D. Knight

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics 3rd Edition by Randall D. Knight

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College Physics by Openstax Chapter 3 Problem 37

Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The baseline is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose service line is 6.40 m from the net?

Solution:

Note: The publication of the solution to this problem is on its way. Sorry for the inconvenience.

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College Physics by Openstax Chapter 3 Problem 36

The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

Solution:

We are required to solve for the maximum distance. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:

The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
The jumper is on level ground, and the motion started from the ground.

The formula for range is

\text{R}=\frac{\text{v}_{\text{o}}^2\sin 2\theta _{\text{o}}}{\text{g}}

Since we are already given the necessary details, we can now solve for the range.

\begin{align*} \text{R}&=\frac{\left(9.5\:\text{m/s}\right)^2\:\sin 90^{\circ} }{9.81\:\text{m/s}^2}\\ \text{R}&=9.20\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

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College Physics by Openstax Chapter 3 Problem 35

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Solution:

We are required to solve for the distance in a standing broad jump. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:

The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
The jumper is on level ground.

The formula for the range is

\text{R}=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}}}{\text{g}}

To find the initial velocity of the jump, vo, we shall use the kinematic formula from the crouch position to the time the person leaves the ground.

\text{v}_{\text{f}}^2=\text{v}_{\text{o}}^2+2\text{ax}

In this case, the final velocity will be the initial velocity of the jump.

\begin{align*} \text{v}_{\text{f}}=\sqrt{\left(0\:\text{m/s}\right)^2+2\left(1.25\times 9.81\:\text{m/s}^2\right)\left(0.600\:\text{m}\right)}=3.84\:\text{m/s} \end{align*}

So, the initial velocity of the flight is 3.84 m/s. We can now use the formula for range.

\begin{align*} \text{R}&=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta_{\text{o}}}{\text{g}} \\ \text{R}&=\frac{\left(3.84\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}\\ \text{R}&=1.50\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 3 Problem 34

An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later.
(a) What is the height of the cliff?
(b) What is the maximum height reached by the arrow along its trajectory?
(c) What is the arrow’s impact speed just before hitting the cliff?

Solution:

Consider the following illustration:

An arrow shot at a height of 1.5 m towards a cliff of height H

Part A

We are required to solve for the value of H. We shall use the formula

\Delta \text{y}=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2

or, we can also write the formula as

\text{y}-\text{y}_0=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2

Substituting the given values, we have

\begin{align*} \text{y}-\text{y}_0 &=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2 \\ \text{H}-1.5\:\text{m} & =\left(30\:\text{m/s}\:\sin 60^{\circ} \right)\left(4.0\:\text{s}\right)+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right)^2 \\ \text{H}-1.5\:\text{m} & =25.44\:\text{m} \\ \text{H} & =25.44\:\text{m}+1.5\:\text{m} \\ \text{H} & =26.94\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Part B

The maximum height of the projectile is given by the formula

\Delta \text{y}=\frac{\text{v}_{\text{0y}}^2}{2\text{g}}

or the formula can be written as

\text{y}_{\text{max}}-\text{y}_{\text{0}}=\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}

Therefore, we have

\begin{align*} \text{y}_{\text{max}}-\text{y}_{\text{0}} & =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}} \\ \text{y}_{\text{max}}& =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}+\text{y}_{\text{0}} \\ \text{y}_{\text{max}}&=\frac{\left(\left(30\:\text{m/s}\right)\sin 60^{\circ} \right)^2}{-2\left(-9.81\:\text{m/s}^2\right)}+1.5\:\text{m} \\ \text{y}_{\text{max}}&=34.40\:\text{m}+1.5\:\text{m} \\ \text{y}_{\text{max}} & =35.90\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

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Part C

To solve for the final speed, we need the vertical and horizontal components when the arrow hits the cliff.

We are going to use the formula for acceleration along the vertical to solve for the final speed in the vertical direction. That is

\text{a}=\frac{\text{v}_{\text{y}}-\text{v}_{\text{0y}}}{\text{t}}

Solving for v_fy in terms of the other variables, we have

\begin{align*} \text{v}_{\text{y}}&=\text{v}_{\text{0y}}+\text{at}\\ \text{v}_{\text{y}}&=\left(30\:\text{m/s}\right)\sin 60^{\circ} +\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right) \\ \text{v}_{\text{y}}&=-13.25\:\text{m/s} \end{align*}

Since we know that the horizontal component of the velocity does not change along the entire flight, we can equate the initial and final horizontal velocities. That is

\begin{align*} \text{v}_{\text{x}}&=\text{v}_{\text{0x}}=\left(30\:\text{m/s}\right)\cos 60^{\circ} \\ \text{v}_{\text{x}}&=15\:\text{m/s} \end{align*}

Therefore, the final speed is

\begin{align*} \text{v}&=\sqrt{\text{v}_{\text{y}}^2+\text{v}_{\text{x}}^2} \\ \text{v}&=\sqrt{\left(-13.25\:\text{m/s}\right)^2+\left(15\:\text{m/s}\right)^2}\\ \text{v}&=20.01\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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