Someone - Ramp Launch And Friction

Click For Summary
SUMMARY

The discussion focuses on a physics problem involving a 2.0 kg wood block launched up a wooden ramp inclined at a 35-degree angle with an initial speed of 10 m/s and a kinetic friction coefficient (μk) of 0.20. The block reaches a vertical height of 2.04 meters above its starting point, calculated using the equations of motion and energy conservation principles. The net force acting on the block is determined by subtracting the frictional force from the launch force, leading to a deceleration that affects the block's ascent and descent.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with energy conservation principles
  • Knowledge of friction and its effects on motion
  • Ability to apply kinematic equations in physics
NEXT STEPS
  • Study the concept of energy conservation in mechanical systems
  • Learn about the effects of friction on inclined planes
  • Explore kinematic equations for motion under constant acceleration
  • Investigate the relationship between kinetic and potential energy
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the dynamics of motion on inclined surfaces.

kraaaaamos
Messages
20
Reaction score
0
Someone Please Help - Ramp Launch And Friction !

1. Homework Statement

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 35* angle. The block’s initial speed is 10m/s. (Given that μk = 0.20)

a. What vertical height does the block reach above its starting point?
b. What speed does it have when it slides back down to its starting point?

2. Homework Equations



3. The Attempt at a Solution

FOR PART A:

I know that initial velocity (y component) is Vi = Vi sin theta
= 10 sin 35
= 5.73m/s
I know we use the equation:
Vf^2 = Vi^2 + 2a (deltaY)

So we need the value of a . . .

Fnet = ma, but we need to knwo the value of Fnet
Fnet = Flaunch - Fk?
We know that Fk = coeff (m)(g)
= 0.2 (2)(-9.8)
= -3.92N
Do we need to find the value of the force of the launch?

I assumed that the value of the launch can be divided into the x-component and the y-component

vi(x) = Vicos theta
= 10 cos 35
= 8.19 m/s
vi(y) = Vi sin theta
= 10 sin 35
= 5.73 m/s

and if we plug in those values to get their overall magnitude

sqrt ( vi(y)^2 + vi(x)^2 )
sqrt [ (8.19)^2 + (5.73)^2 ]
sqrt [ 67.1 + 32.9]
sqrt (100)
= 10

Flaunch = 10 ( mass)
= 10 (2)
= 20N

Fnet = 20 - 3.92
= 16.08

Fnet = ma
16.08 = (2)a
a = 8.04 m/s2

Vf(y)^2 = Vi(y)^2 + 2(ay)(deltaY)
0^2 = 5.73^2 + 2 (-8.04)(deltay)
-32.8 = -16.08 (deltaY)
deltaY = 2.04m <<< FINAL ANSWER...

is that corect? Everything from the point where I suggested that Fnet = Flaunch - Fk . . . was something I deduced on my own. So I have no idea if it even makes sense.
 
Physics news on Phys.org
Please do not post a problem more than once.


One does not need to know the force at launch, but only that the speed is 10 m/s up the ramp. One does not need x,y of velocity.

The 2 kg mass has an initial velocity up the ramp. Without friction is would decelerate under the influence of gravity and reach a certain height where it stops. The change in KE (1/2mvo2 - 0) would equal the change in gravitational potential energy, mgh, where h is the change in height (elevation) in the gravity field.

With friction there is a constant dissipation of energy. So in addition to the deceleration under the component of gravity pointing down the incline, there is also a deceleration component related to friction.

In the opposite direction the mass slides down the ramp with friction pointing up the ramp.

Here is a good reference on friction on block sliding on an incline.
http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html#c2
 
Last edited:

Similar threads

  • · Replies 2 ·
Replies
2
Views
6K
Replies
18
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
4
Views
4K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K