EK Physics 1001 #358

[PhaZed]

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Hello everyone,

I was wondering if you could help me with the following question from the EK 1001 Physics book.

Question #358--there is a figure.

There is a frictionless curved surface (a quarterpipe) with a 500 gram block resting on it. The block is 4 cm off the ground. The block starts from rest and slides down the curve. At the bottom of the curve, there is a flat portion with a kinetic friction coefficient between the block and flat area of 0.2. How far does the block travel along the flat portion of the ramp?

It states that the answer is 20cm, however I am a little unclear of how they got that answer. I understand that Potential energy is converted to kinetic energy as it is falling, and friction opposes the kinetic energy. However, I am unsure of how to set up the equation.

Thanks

 

[ilovemcat]

Hello everyone,

I was wondering if you could help me with the following question from the EK 1001 Physics book.

Question #358--there is a figure.

There is a frictionless curved surface (a quarterpipe) with a 500 gram block resting on it. The block is 4 cm off the ground. The block starts from rest and slides down the curve. At the bottom of the curve, there is a flat portion with a kinetic friction coefficient between the block and flat area of 0.2. How far does the block travel along the flat portion of the ramp?

It states that the answer is 20cm, however I am a little unclear of how they got that answer. I understand that Potential energy is converted to kinetic energy as it is falling, and friction opposes the kinetic energy. However, I am unsure of how to set up the equation.

Thanks

Energy starts off as potential energy. As the block slides, potential energy is transfered into kinetic energy.
At the bottom of the ramp (curve), kinetic energy is at it's maximum (PE is 0J):

mgh = (0.5kg)(10m/s^2)(4x10^-2m)
mgh = 0.2 J

mgh = KE, therefore KE = 0.2J

Fricton = mkN
Friction = (0.2)(5)
Friction = 1N

Using the Work-Energy Theorem:
deltaKE=Fd where F represents friction.

deltaKE = Fd
0.2J = 1Nd
d = 0.2m or 20cm (choice B)

 

[PhaZed]

Thank you for your response. It made perfect sense. However, how would you perform the problem if the block had an initial velocity of 2 m/s?

The next question is states: "The block has a mass of 50g and h is 20 cm. The block has an initial velocity of 2m/s, and slides down the curve. If the coefficient of kinetic friction between the block and the flat portion of the ramp is 0.2, how far does the block travel along the flat portion of the ramp?"

I tried starting it out the same way as the previous question, but got 1m instead of the the correct answer (which is 2m).

At the end of the trip, the block has all KE, so I did the following:

KE: 1/2 mv^2
KE: 1/2(0.05)(2^2)
KE: 0.1J

However, I am not sure if this is right. I am not sure how having an initial velocity effects the equation.

Anyway, with the above KE, I calculated the frictional force,

Fk= (uk)(m)(g)
Fk= (0.2)(0.05)(10)
Fk= 0.1 N

DeltaKE = Fd
0.1J = 0.1d
d= 1m?

What did I do wrong here?

Thanks

 

[ilovemcat]

Thank you for your response. It made perfect sense. However, how would you perform the problem if the block had an initial velocity of 2 m/s?

The next question is states: "The block has a mass of 50g and h is 20 cm. The block has an initial velocity of 2m/s, and slides down the curve. If the coefficient of kinetic friction between the block and the flat portion of the ramp is 0.2, how far does the block travel along the flat portion of the ramp?"

I tried starting it out the same way as the previous question, but got 1m instead of the the correct answer (which is 2m).

At the end of the trip, the block has all KE, so I did the following:

KE: 1/2 mv^2
KE: 1/2(0.05)(2^2)
KE: 0.1J

However, I am not sure if this is right. I am not sure how having an initial velocity effects the equation.

Anyway, with the above KE, I calculated the frictional force,

Fk= (uk)(m)(g)
Fk= (0.2)(0.05)(10)
Fk= 0.1 N

DeltaKE = Fd
0.1J = 0.1d
d= 1m?

What did I do wrong here?

Thanks

I'm pretty sure that's an error by EK as I got the same answer:

PEi + KEi = PEf + KEf (where PEf = 0J because you're at the surface)
mgh + 1/2mvi^2 = KEf
(0.05kg)(10m/s^2)(0.2m) + 1/2(0.05kg)(2m/s)^2 = KEf
0.1 J = KEf

Friction: mkN
Friction: (0.2)(0.5N)
Friction: 0.1N

KE = Fd
0.1J = 0.1N x distance
0.1J/0.1N = distance
distance = 1m

 

[ilovemcat]

I'm pretty sure that's an error by EK as I got the same answer:

PEi + KEi = PEf + KEf (where PEf = 0J because you're at the surface)
mgh + 1/2mvi^2 = KEf
(0.05kg)(10m/s^2)(0.2m) + 1/2(0.05kg)(2m/s)^2 = KEf
0.1 J + 0.1 J= KEf
0.2 J = KEf

Friction: mkN
Friction: (0.2)(0.5N)
Friction: 0.1N

KE = Fd
0.2J = 0.1N x distance
0.2J/0.1N = distance
distance = 2m

Changes in RED

Oops, looks like I made the same mistake you did 🙂 I forgot to add 0.1 J twice, so it should say 0.2 J = KEf.

0.2J/0.1N = 2m, so yeah they are correct. I don't know why they insist on using complicated numbers.