Conservation of Energy speed of block

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Homework Help Overview

The discussion revolves around a block sliding down a frictionless track with a rough section, analyzing its speed as it leaves the track and its subsequent motion through the air until it hits the ground. The problem involves concepts of gravitational potential energy, kinetic energy, and the effects of friction.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy principles, questioning how to incorporate the effects of friction into the calculations for velocity. There is mention of using the work-energy theorem to account for energy lost due to friction.

Discussion Status

Some participants are exploring the relationship between kinetic energy and potential energy while considering the impact of friction. There is a request for clarification on how heat energy relates to the problem, indicating a lack of consensus on the role of friction in the calculations.

Contextual Notes

Participants note that heat energy has not been discussed in their class, which may affect their understanding of the problem. The original poster expresses uncertainty about how to account for friction in their calculations.

himetx
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A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track (as shown in the figure attached for#007). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.8 m/s2 .

m=0.405 kg
H=5.2 m
h=2 m
rough area=1.2m
μ=0.3

A. What is the speed v of the block when it leaves the track? Answer in units of m/s.

B. What is the horizontal distance x the block travels in the air? Answer in units of m.

C. What is the total speed of the block when it hits the ground? Answer in units of m/s.


The initial height of the block when it is at rest at the top of frictionless track = H

Height of the point where the block leaves the track horizontally = h

Change in height = H - h

Loss in gravitational potential energy of block=mg(H-h)

Gain in kinetic energy of block as it leaves the track =Loss in gravitational potential energy of block

Gain in kinetic energy =(1/2)mv^2

(1/2)mu^2=mg(H - h)

u = sq rt 2g (H - h)

PART A. The ball leaves the track with a speed = sq rt 2g ( H - h)

If H and h are in meter then ball leaves the track with a speed = sq rt19.62 (H-h) m/s

I understand how to get to this point, but I don't know how to account for the friction that needs to be considered to determine the velocity.

PART C. If V is speed with which block hits the ground.

Final kinetic energy of block (KE) =initial potential energy of block (PE)

(1/2)mV^2=mgH

V= sq rt 2gH

the speed of the block when it hits the ground is sq rt 2gH
If H is in meter then speed is [ sq rt 19.62 H ] m/s
 

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himetx said:
I understand how to get to this point, but I don't know how to account for the friction that needs to be considered to determine the velocity.

The conservation law of energy states that energy is never created or destroy, only converted.

When there is friction between two objects, there is a loss in kinetic energy, but this energy doesn't disappear. It's converted into heat (and sound, and sometimes other forms of energy. But only heat here).

KE_{1} + PE_{1} = KE_{2} + PE_{2} + E_{h}

Heat energy caused by friction is simply determined by:

E_{h} = F_{f}d = \mu mgd

If that doesn't help, let me know.
 
I don't think heat energy applies to this problem. We have yet to discuss heat energy in my class. If you can show me how this applies using the values given, that would help me understand what your trying to convey.
 
welcome to pf!

hi himetx! welcome to pf! :wink:
himetx said:
I understand how to get to this point, but I don't know how to account for the friction that needs to be considered to determine the velocity.

you can use conservation of energy, but for the rough patch you need to use the work energy theorem: https://www.physicsforums.com/library.php?do=view_item&itemid=75" = change in (mechanical) energy …

so find the work done by https://www.physicsforums.com/library.php?do=view_item&itemid=39", and subtract it from the energy :smile:
 
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