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Bigflakes
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#1
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#1
A square block of mass 10 grams rests on a rough plane which is inclined to the horizontal at an
angle of 8 degrees. A force of 0.03 N, acting in a direction parallel to the line of greatest slope, is
applied so as to move it up the plane at a constant velocity. When the block has travelled a
distance of 1.1 m from its initial position the applied force is removed. The block moves on and
comes to rest after travelling a further 0.25 m.
Calculate:
i) the work done by the applied force.
) the total gain in potential energy of the block.
li) the work done against the frictional force.
iv) the frictional force.
v) the normal reaction of the block.

Can anyone help with any of these?
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Callicious
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#2
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#2
(Original post by Bigflakes)
A square block of mass 10 grams rests on a rough plane which is inclined to the horizontal at an
angle of 8 degrees. A force of 0.03 N, acting in a direction parallel to the line of greatest slope, is
applied so as to move it up the plane at a constant velocity. When the block has travelled a
distance of 1.1 m from its initial position the applied force is removed. The block moves on and
comes to rest after travelling a further 0.25 m.
Calculate:
i) the work done by the applied force.
) the total gain in potential energy of the block.
li) the work done against the frictional force.
iv) the frictional force.
v) the normal reaction of the block.

Can anyone help with any of these?
To get purely via SUVAT/force resolutions and not considering energy theorems or anything else to get friction:
Draw a free-body diagram (draw the plane on, too- it helps a lot.) Label the forces when you do and do this for the initial state of the system (i.e. when the applied force is being... applied.) Note: "Move it up the plane at constant velocity." Think of the forces on your diagram- where is the work by the applied force going if not to accelerate? There are two sinks (one of which is non-conservative.) Resolve the forces parallel and orthogonal, using symbols!!! There shouldn't be a single number in your force resolution.

Once the block stops it decelerates and the applied force is gone- redraw the diagram for this point and re-resolve forces. Once you've done the two free-body diagrams and resolved forces parallel and orthogonal to the plane, using symbols, post your working here (alternatively someone more benevolent might provide more of a solution.)

To solve this kind of thing you need to know:
- Use of free-body diagrams
- Force resolution with inclined planes
- Frictional forces (namely kinetic friction)
- Energy conservation (and lack of)
Last edited by Callicious; 7 months ago
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Callicious
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#3
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#3
To get you started on the symbols bit-



\begin{align}

\theta = 8^\circ\\ 

m = 0.010 \textrm{kg} \\ 

F_{||} = 0.03 \textrm{N} \\ 

u = u \\ 

v = 0 \\ 

\Delta{x_{||,1}} = 1.1 \textrm{m} \\

\Delta{x_{||,2}} = 0.25 \textrm{m} \\ 

g = g 

\end{align}

That kind of thing. That's just how I would do it though. I was taught to use a lowercase f for net forces hence why I opted for the big F for applied.
Last edited by Callicious; 7 months ago
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Bigflakes
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#4
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#4
(Original post by Callicious)
Draw a free-body diagram (draw the plane on, too- it helps a lot.) Label the forces when you do and do this for the initial state of the system (i.e. when the applied force is being... applied.) Note: "Move it up the plane at constant velocity." Think of the forces on your diagram- where is the work by the applied force going if not to accelerate? There are two sinks (one of which is non-conservative.) Resolve the forces parallel and orthogonal, using symbols!!! There shouldn't be a single number in your force resolution.

Once the block stops it decelerates and the applied force is gone- redraw the diagram for this point and re-resolve forces. Once you've done the two free-body diagrams and resolved forces parallel and orthogonal to the plane, using symbols, post your working here (alternatively someone more benevolent might provide more of a solution.)

To solve this kind of thing you need to know:
- Use of free-body diagrams
- Force resolution with inclined planes
- Frictional forces (namely kinetic friction)
- Energy conservation (and lack of)
Thank you, for ii) would I use trig to work out how highly it has moved and then just use mgh
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Callicious
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#5
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#5
(Original post by Bigflakes)
Thank you, for ii) would I use trig to work out how highly it has moved and then just use mgh
Yup yup. Decompose the gain parallel to the slope into a \hat{y} component (what you want for mgh.)
Last edited by Callicious; 7 months ago
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Bigflakes
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#6
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#6
(Original post by Callicious)
Yup yup. Decompose the gain parallel to the slope into a \hat{y} component (what you want for mgh.)
Okay and for iv) wouldn’t the frictional force be the same as 0.03N as it’s moving at constant velocity
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Callicious
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#7
(Original post by Bigflakes)
Okay and for iv) wouldn’t the frictional force be the same as 0.03N as it’s moving at constant velocity
"A force of 0.03 N, acting in a direction parallel to the line of greatest slope, is applied so as to move it up the plane at a constant velocity"

Look at your force resolution parallel to the plane. There should be something along the lines of
m\ddot{x_{||}} = \sum_i F_{||,i}
On the right side, you should have three terms- weight, friction, and the applied force, all with signs afforded to them (\pm.) If the block isn't accelerating, this is clearly equal to zero: is friction really equal to the applied force?
Last edited by Callicious; 7 months ago
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Bigflakes
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#8
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#8
(Original post by Callicious)
"A force of 0.03 N, acting in a direction parallel to the line of greatest slope, is applied so as to move it up the plane at a constant velocity"

Look at your force resolution parallel to the plane. There should be something along the lines of
m\ddot{x_{||}} = \sum_i F_{||,i}
On the right side, you should have three terms- weight, friction, and the applied force, all with signs afforded to them (\pm.) If the block isn't accelerating, this is clearly equal to zero: is friction really equal to the applied force?
I’m sorry but I’m really confused, is it asking for the frictional force as the block is moving or when it came to a stop?
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papie
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#9
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#9
(Original post by Bigflakes)
I’m sorry but I’m really confused, is it asking for the frictional force as the block is moving or when it came to a stop?
I think the frictional force remains constant. BTW you can also use the work energy theorem. Will make this type of problem easier to solve.
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Callicious
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#10
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#10
(Original post by Bigflakes)
I’m sorry but I’m really confused, is it asking for the frictional force as the block is moving or when it came to a stop?
(Original post by papie)
I think the frictional force remains constant. BTW you can also use the work energy theorem. Will make this type of problem easier to solve.
Frictional force is kinetic in both cases \therefore the same. I don't remember the technical terms for this kind of thing but if it's energy conservation you're looking to apply, watch out for non-conservation by kinetic friction.

If you write a force resolution down it definitely helps you think the problem through a bit more rather than working it through mentally entirely (you can do this easily with practice.)

In terms of energy, though... (Work Applied) = (Gravity Potential + Friction Work)

You should be able to work that to get a value for the friction coefficient/etc, as opposed to using the deceleration to work that out instead like I initially posited.

Uwa- I didn't realize you weren't the OP when I replied! Sorry :lol:
Last edited by Callicious; 7 months ago
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