How Far Will a Block Slide Up an Incline Before Stopping?

In summary, the block with mass 23.4 kg slid down an inclined plane of slope angle 31.1 degrees with a constant velocity, and was then projected up the same plane with an initial speed of 1.50 m/s. The frictional force was found by using the angle and the component of mg along the plane, resulting in a total of -2mgsin(theta). To find the distance traveled up the incline, the initial speed, acceleration, and final velocity of 0 were used in kinematics equations.
  • #1
bulldog23
120
0

Homework Statement



A block with mass m = 23.4 kg slides down an inclined plane of slope angle 31.1 o with a constant velocity. It is then projected up the same plane with an initial speed 1.50 m/s. How far up the incline will the block move before coming to rest?



Homework Equations


KE=1/2mv^2
W=F*displacement*cos(theta)


The Attempt at a Solution


I don't really know how to even start this problem. Can someone walk me through it?
 
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  • #2
Use the coming downward part to find out the frictional force. There are different ways of doing this...

one ways is: work by friction = change in mechanical energy.

once you know the frictional force, you can do the next part.
 
  • #3
I haven't ever done anything with mechanical energy. How do you figure that out?
 
  • #4
bulldog23 said:
I haven't ever done anything with mechanical energy. How do you figure that out?

hmmm... actually I was wrong... that's probably not the best way to do the first part... probably best to just use forces...

The net force acting on the block as it slides downwards is 0 (since acceleration is constant). Can you use this to figure out the frictional force?
 
  • #5
so then frictional force must be equal but opposite to mg??
 
  • #6
bulldog23 said:
so then frictional force must be equal to mg??

not mg... what is the component of mg along the plane?
 
  • #7
I'm not quite sure. Is it just g?
 
  • #8
bulldog23 said:
I'm not quite sure. Is it just g?

no... mg divides into two components... one perpendicular to the plane... one parallel to the plane.

use the angle 31 degrees.
 
  • #9
m*cos(theta)
 
  • #10
bulldog23 said:
m*cos(theta)

no. what you need is mgsin(theta).
 
  • #11
alright, my bad I meant to put mgcos(theta), but I see why it's mgsin(theta). So then I figure that out and divide by 1.5 m/s?
 
  • #12
bulldog23 said:
alright, my bad I meant to put mgcos(theta), but I see why it's mgsin(theta). So then I figure that out and divide by 1.5 m/s?

well... mgsin(theta) is the frictional force... the magnitude doesn't change when going up or down... but the direction is different...

So when the object is going up, what is the net force acting on it along the plane?

Take the net force... divide by m... that's the acceleration.
 
  • #13
when going up would it be -mgsin(theta)?
 
  • #14
bulldog23 said:
when going up would it be -mgsin(theta)?

yes, that's the frictional force... but you've also got -mgsin(theta) due to gravity... so it's a total of -2mgsin(theta).
 
  • #15
so when I plug the numbers in I get -6.74 m. Is this right?
 
  • #16
bulldog23 said:
so when I plug the numbers in I get -6.74 m. Is this right?

I don't think so. How did you get that?
 
  • #17
-2mgsin(theta)/m
 
  • #18
bulldog23 said:
-2mgsin(theta)/m

so that's -2gsin(theta). I get -10.12m/s^2
 
  • #19
oh ok. So then I divide by the initial speed to get how far up it travels?
 
Last edited:
  • #20
bulldog23 said:
oh ok. So then I divide by the initial speed?

No. You need to get the distance... you have v0, a and vf = 0. use kinematics to get the distance.
 
  • #21
alright, thanks for the help!
 
  • #22
bulldog23 said:
alright, thanks for the help!

Did you get the distance?
 

1. What is a block on an inclined plane?

A block on an inclined plane is a physics problem where a solid object, typically a block, is placed on a flat surface that is tilted at an angle. The block is subject to two main forces: gravity pulling it downwards and the normal force from the surface pushing it upwards.

2. What factors affect the motion of a block on an inclined plane?

The motion of a block on an inclined plane is affected by the angle of the incline, the mass of the block, and the coefficient of friction between the block and the surface. These factors determine the magnitude and direction of the forces acting on the block.

3. How do you calculate the net force on a block on an inclined plane?

To calculate the net force on a block on an inclined plane, you can use the formula Fnet = mg sinθ - μmg cosθ, where Fnet is the net force, m is the mass of the block, g is the acceleration due to gravity, θ is the angle of the incline, and μ is the coefficient of friction.

4. What is the relationship between the angle of the incline and the motion of the block?

The angle of the incline affects the motion of the block by changing the components of the gravitational force acting on it. As the angle increases, the component of the force pulling the block down the incline increases, causing the block to accelerate faster. At a certain angle, the block will begin to slide down the incline due to the force of gravity being greater than the frictional force.

5. How do you determine the acceleration of a block on an inclined plane?

The acceleration of a block on an inclined plane can be calculated using the formula a = gsinq - μgcosq, where a is the acceleration, g is the acceleration due to gravity, q is the angle of the incline, and μ is the coefficient of friction. This formula takes into account the effects of both gravity and friction on the block's motion.

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