# SUVAT Problem: Box moving on inclined plane

• dorothy
In summary, the conversation is discussing a physics problem involving a box being projected horizontally and landing on an inclined plane. The conversation addresses the question of how far up the plane the box goes and its velocity at that point. The participants also discuss the correct units and conventions for acceleration and velocity in this situation.
dorothy
Homework Statement
Does anyone knows how to solve these questions? Thanks a lot!
Relevant Equations
v=u+at
s=ut+1/2at^2
v^2=u^2+2as

Last edited:
You seem to already have solved everything except (c), for which you have not shown any effort. What are your own thoughts regarding (c)?

dorothy said:
Homework Statement:: Does anyone knows how to solve these questions? Thanks a lot!
Relevant Equations:: v=u+at
s=ut+1/2at^2
v^2=u^2+2as

View attachment 299486
In part c, how far up the plane does the box go? What is its velocity at this point?

Orodruin said:
You seem to already have solved everything except (c), for which you have not shown any effort. What are your own thoughts regarding (c)?
Hi, I want to whether my answers on a & b are correct or not? For (c), I don’t really get the question, like what is the point of projection? How’s the deceleration and acceleration do in this case (c)? Thank you.

Chestermiller said:
In part c, how far up the plane does the box go? What is its velocity at this point?
Before that, I want to ask whether I understand question8 correctly? Is it true that there is a guy throwing(projecting) a box horizontally and the box finally land on the inclined plane(like what I have drawn)? Thank you.

dorothy said:
Hi, I want to whether my answers on a & b are correct or not?
I don't agree with your answer to a). Acceleration is a vector, not a scalar.

This also affects the answer I would give to b) i) I).

Thaty said, it's not clear that the question setter acknowledges that acceleration is a vector.

dorothy
PeroK said:
I don't agree with your answer to a). Acceleration is a vector, not a scalar.

This also affects the answer I would give to b) i) I).

Thaty said, it's not clear that the question setter acknowledges that acceleration is a vector.
Should (a) be -4.905?

dorothy said:
Should (a) be -4.905?
That's what I would put. With the units as you have in your answer.

dorothy
PS It does say velocity ##5 \ m/s## up the incline, which infers a convention that up the incline is positive. I'd prefer, however, that the question setter be more explicit about this. E.g. say a velocity of ##+5 \ m/s## up the incline. Then there is no doubt.

PeroK said:
That's what I would put. With the units as you have in your answer.
I see. What about (b)? Are all of them correct?

dorothy said:
I see. What about (b)? Are all of them correct?
I don't agree with b i) I).

dorothy said:
Before that, I want to ask whether I understand question8 correctly? Is it true that there is a guy throwing(projecting) a box horizontally and the box finally land on the inclined plane(like what I have drawn)? Thank you.
View attachment 299489
I expect the exercise composer means ##5## m/s along the incline, not ##5## m/s in a horizontal diertion.

##\ ##

Chestermiller and PeroK
If we look at part c) it's clear that up is positive. Note, however, that the question setter gets confused with motion down the slope and changes the convention. The acceleration down the slope should be ##-2 \ ms^{-2}##.

This seems to be a common cause of confusion: that negative acceleration is reducing speed (deceleration) and positive acceleration is increasing speed. But, if you think about it, this is not correct. Once we have established that up is positive, then the acceleration of gravity is ##-9.8 \ ms^{-2}## throughout projectile motion. It doesn't change to ##+9.8 \ ms^{-2}## for the descent. And, in fact, the SUVAT equation would no longer work if you change your convention half way through.

In this case, the acceleration does change so you have to reset your SUVAT equations. But, I suggest there is no need to change convention and make down positive.

PeroK said:
I don't agree with b i) I).

May I know why bi) is not correct? I don’t know how to do it.

dorothy said:
May I know why bi) is not correct? I don’t know how to do it.
Sorry I meant b ii) I.

The acceleration has a greater magnitude in this case.

PeroK said:
Sorry I meant b ii) I.

The acceleration has a greater magnitude in this case.
So higher acceleration gives smaller distance traveled until it turns around ... as OP stated.

PeroK said:
Sorry I meant b ii) I.

The acceleration has a greater magnitude in this case.
But if i assume the theta is 45° (which is greater than 30°, tilited more), then i put it into -9.81sin45°. I get the new acceleration=-6.9ms^-2. Next, I put 6.9 into the equation and get the new max distance =1.8 which is smaller than the original 2.55

dorothy said:
But if i assume the theta is 45° (which is greater than 30°, tilited more), then i put it into -9.81sin45°. I get the new acceleration=-6.9ms^-2. Next, I put 6.9 into the equation and get the new max distance =1.8 which is smaller than the original 2.55
That right. It's like having more powerful brakes: you stop in a shorter distance.

Orodruin said:
So higher acceleration gives smaller distance traveled until it turns around ... as OP stated.
So if the plane is tilited more, it becomes steeper, the distance traveled will also become smaller. Does it mean that I am actually correct on bii)1)?

PeroK said:
That right. It's like having more powerful brakes: you stop in a shorter distance.
Does it mean that bii) I) is correct ?

dorothy said:
Does it mean that bii) I) is correct ?

PeroK said:
No worries :) For part c, can you explain to me what is the meaning of reaching the point of projection? I don’t know which point it is. I’m super confused now

dorothy said:
No worries :) For part c, can you explain to me what is the meaning of reaching the point of projection? I don’t know which point it is.
It was my fault.

I can only imagine that it means back at the bottom of the incline!

dorothy said:
No worries :) For part c, can you explain to me what is the meaning of reaching the point of projection? I don’t know which point it is. I’m super confused now
Probably the point where it was released with the original speed, but on the way down.

PeroK said:
It was my fault.

I can only imagine that it means back at the bottom of the incline!

Thank you for your help! :)

## 1. What is the SUVAT problem?

The SUVAT problem is a type of physics problem that involves a box or object moving on an inclined plane. It requires the use of the SUVAT equations, which are a set of equations that relate an object's initial and final velocities, acceleration, and displacement.

## 2. How do you solve a SUVAT problem?

To solve a SUVAT problem, you first need to identify the known and unknown variables. Then, you can use the appropriate SUVAT equation to solve for the unknown variable. It is important to use the correct equation and to pay attention to units and directions when plugging in values.

## 3. What are the SUVAT equations?

The SUVAT equations are five equations that relate an object's initial and final velocities (u and v), acceleration (a), and displacement (s). They are: v = u + at, s = ut + 1/2at^2, v^2 = u^2 + 2as, s = 1/2(u+v)t, and s = vt - 1/2at^2. These equations can be used to solve for any of the five variables, as long as the other four are known.

## 4. How do you apply the SUVAT equations to a box on an inclined plane?

To apply the SUVAT equations to a box on an inclined plane, you must first draw a free body diagram to identify the forces acting on the box. Then, you can use trigonometry to determine the components of the weight force and the normal force. These components can then be used in the SUVAT equations to solve for the acceleration and other variables.

## 5. What are some common mistakes when solving SUVAT problems?

Some common mistakes when solving SUVAT problems include using the wrong equation, not paying attention to units and directions, and not drawing a free body diagram to properly identify the forces acting on the object. It is also important to use consistent units throughout the problem and to round to the appropriate number of significant figures in the final answer.

Replies
5
Views
1K
Replies
44
Views
3K
Replies
8
Views
1K
Replies
5
Views
1K
Replies
19
Views
2K
Replies
9
Views
546
Replies
1
Views
1K
Replies
41
Views
1K
Replies
7
Views
2K
Replies
45
Views
3K