John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of Particle projected up a rough inclined plane

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John Wick 5 years, 11 months ago

Created this as a copy of Simon's copy of Particle projected up a rough inclined plane.

Name	Status	Author	Last Modified
Particle projected up a rough inclined plane	Ready to use	Amy Chadwick	04/08/2016 16:07
John's copy of Particle projected up a rough inclined plane	draft	John Bridges	19/09/2016 00:12
Agus's copy of Particle projected up a rough inclined plane	draft	Agus Suroso	14/01/2019 08:25
Simon's copy of Particle projected up a rough inclined plane	draft	Simon Thomas	29/04/2019 10:21
John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of Particle projected up a rough inclined plane	draft	John Wick	29/04/2019 10:32
plano inclinado	draft	Ivan Munoz	07/12/2022 22:18

There is one other version that you do not have access to.

Question statement

A particle of mass $\var{mass} \mathrm{kg}$ is projected with speed $\var{u}\ \mathrm{ms^{-1}}$ up a rough plane. The coefficient of friction between the particle and the plane is $\var{mu}$ . The plane is inclined at an angle $\theta = \var{theta}^{\circ}$ to the horizontal.

The acceleration due to gravity is $g = 9.8 \mathrm{ms^{-2}}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

g	number	9.8
mu	number	0.36
theta	number	58.5
u	integer	15
mass	integer	9
R	number	46.0843734079
a	number	-10.199248547
s	number	11.0302243819

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
9.8

JME function reference

Description

The acceleration due to gravity.

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `number`

9.8

→ Used by:

Parts

Part a) Gap-fill
1. Gap Number entry
  Add
Part b) Number entry
Part c) Number entry
Part d) Number entry

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

What is the normal reaction, $R$ , between the particle and the plane?

Give your answer in Newtons to 3 decimal places.

$R =$ Unnamed gap

Advice

A diagram can be drawn to show the forces acting on the particle.

Here the particle is drawn in three positions, showing its original speed when it is projected from the bottom of the slope, its position at some point up the slope and its position when it instantaneously comes to rest further up the slope.

The slope is rough so friction ( $\mu R$ N) acts down the slope, against the direction of motion.

a)

The normal reaction $R$ , is found by resolving the forces perpendicular to the plane.

$\begin{align} R - mg \cos \theta & = 0, \\ R & = mg \cos \theta, \\ & = (\var{mass} \times 9.8) \cos (\var{theta}^{\circ}), \\ &= \var{precround(R,3)} \mathrm{N}. \end{align}$

The normal reaction force between the particle and the plane is $\var{precround(R,3)} \mathrm{N}$ .

b)

To find the acceleration of the particle we resolve the forces parallel to the plane.

$\begin{align} - mg \sin \theta - \mu R & = ma, \\ - \var{mass} \times 9.8 \sin (\var{theta}^{\circ}) - (\var{precround(R,3)} \times \var{mu}) & = \var{mass}a, \\ a & = \frac{ - \var{mass} \times 9.8 \sin (\var{theta}^{\circ}) - (\var{precround(R,3)} \times \var{mu}) }{\var{mass}}, \\ & = \var{precround(a,3)} \mathrm{ms^{-2}}. \end{align}$

Therefore the deceleration of the particle is $\var{precround(-a,3)} \mathrm{ms^{-2}}$ as it is the negative of the acceleration.

c)

The particle will travel up the slope until it comes to instantaneous rest; at this point its velocity will be $0$ . We can use the SUVAT equation $v^2 = u^2 + 2as$ in order to find the distance the particle will travel.

We know that $u = \var{u}, v= 0$ and $a = \var{precround(a,3)}$ .

$\begin{align} v^2 & = u^2 + 2as, \\ 0 & = \simplify{{u}^2+{precround(2a,3)}s}, \\ s & = \frac{\var{u}^2}{\var{precround(-2a,3)}}, \\[0.5em] & = \var{precround(u^2/(-2*a),3)} \mathrm{m}. \end{align}$

The distance the particle will travel up the plane before it comes to instantaneous rest is $\var{precround(u^2/(-2*a),3)} \mathrm{m}$ .

Give a worked solution to the whole question.

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John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of Particle projected up a rough inclined plane

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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John Wick 5 years, 11 months ago

Error in variable testing condition

Error

Warning

Generated value: `number`

Parts

Gap-fill

a)

b)

c)

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John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of Particle projected up a rough inclined plane

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

John Wick 5 years, 11 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

a)

b)

c)

Generated value: `number`