How much work, speed and distance

[Haveagoodday]

Homework Statement

A child wants to go ride on her sleigh, and she decides to walk up the Big

Hill, of height 100 m relative to where she starts. There are two ways up;

the steep way, where the slope is 30.0 degrees, and the not-so steep way, where the

slope is 15.0 degrees. The coefficient of kinetic friction between snow and sleigh is

fs k = 0,100. The mass of the sleigh is 8.00 kg, the mass of the child is 30.0kg.

a) How much work does the child have to do on the sleigh to get it to the

top of the hill: the steep way? The not-so-steep way? How much potential

energy does the sleigh now have relative to the starting point?

b) The child now sits down on the sleigh and slides down the steep side.

What is her speed as she gets to the bottom of the hill? What if she would

have gone down the not-so-steep side?

c) Assuming that after reaching the bottom, she can continue on a horizontal

surface of snow, how far does she slide before coming to a halt (steep and

not-so-steep)?

The Attempt at a Solution

a) In problem a i got
The amount of work done on the sleigh the steep way = 2425 J
The amount of work done on the sleigh the not steep way = 10813 J
Potential energy=7840 J

b) Speed down steep side= 40,3 m/s
Speed down not steep side= 35 m/s

c) I haven't solved c yet.

All answers are appreciated!
[/B]

 

[stockzahn]

a) In problem a i got
The amount of work done on the sleigh the steep way = 2425 J
The amount of work done on the sleigh the not steep way = 10813 J
Potential energy=7840 J

The potential energy the sleigh gained is larger than the work the girl did on it - that can't be possible (in the steep way case).

b) Speed down steep side= 40,3 m/s
Speed down not steep side= 35 m/s

I got 39.6 m/s and 34.7 m/s (probably due to different values of g), so I agree.

c) I haven't solved c yet.

You can solve it with energy consistency: you know the kinetic energy and can convert it into work done by the friction of the snow.

 

[Haveagoodday]

The potential energy the sleigh gained is larger than the work the girl did on it - that can't be possible (in the steep way case).
I got 39.6 m/s and 34.7 m/s (probably due to different values of g), so I agree.
You can solve it with energy consistency: you know the kinetic energy and can convert it into work done by the friction of the snow.

This is the method that i used to calculate the work on steep way:
Fx= Fsin30-μn=0
Fy= n-mgcos30=0
n=mgcos30
Fsin30-μ(mgcos30)=0
F=μ(mgcos30)/sin30=14N

L= distance from standpoint till top
L= 100m/sin30=200m

W=F*rcosθ
W=14N*200m*cos30= 2425J

the same method used with the not steep way, but with different θ and L

for the potential energy:

U=mgy=8kg*9.80m/s^2*100m=7840J

 

[tom ryen]

in b) i did:

Steep way:
m=38 , g=9.8, h=100 , fk=6.789, d = 200
vf= sqrt((2/m)*(mgh-fkd))= 43.46 m/s

not so steep way:
m=38 , g=9.8, h=100 , fk=7.57, d = 386.37
vf= sqrt((2/m)*(mgh-fkd))= -> 42.49 m/s

 

[stockzahn]

This is the method that i used to calculate the work on steep way:
Fx= Fsin30-μn=0
Fy= n-mgcos30=0
n=mgcos30
Fsin30-μ(mgcos30)=0
F=μ(mgcos30)/sin30=14N

If I don't misread your post, you placed the x-axis of the coordinate system parallel to the slope and y-axis perpendicular to it.

For ΣF_x:

F_f = μ ⋅ N ... friction force
F ... girl's pulling force → why do you multiply it with the sine of the hill's angle?
plus: you forgot, that the weight has a component in x-direction to

ΣF_y looks good to me

L= distance from standpoint till top
L= 100m/sin30=200m

W=F*rcosθ
W=14N*200m*cos30= 2425J

the same method used with the not steep way, but with different θ and L

The length of the slope L is calculated correctly.

When calculating the work why do you multiply it with the cosine of θ? The girl is pulling the sleigh parallel to the slope, you already calculated the length L, which you need for the calculation of the work.

1) Think about the sum of the forces in x-direction again:
a) direction of the girl's pulling force (or at least the component that is able to do work)
b) the component of the weight in x-direction

2) To calculate the work multiply the pulling force with the length the sleigh is actually pulled (the whole slope)

U=mgy=8kg*9.80m/s^2*100m=7840J

That is correct and it must be less than the work of the girl, else you create a perpetuum mobile.

 

[stockzahn]

in b) i did:

Steep way:
m=38 , g=9.8, h=100 , fk=6.789, d = 200
vf= sqrt((2/m)*(mgh-fkd))= 43.46 m/s

not so steep way:
m=38 , g=9.8, h=100 , fk=7.57, d = 386.37
vf= sqrt((2/m)*(mgh-fkd))= -> 42.49 m/s

For the friction force I get: F_f = N⋅μ = m⋅g⋅cosθ⋅μ
θ = 30° → F_f = 32.3 N
θ = 15° → F_f = 36.0 N

 

[tom ryen]

For the friction force I get: F_f = N⋅μ = m⋅g⋅cosθ⋅μ
θ = 30° → F_f = 32.3 N
θ = 15° → F_f = 36.0 N

You are right, i used m=8 on my calculator instead of m=38.

 

[Haveagoodday]

If I don't misread your post, you placed the x-axis of the coordinate system parallel to the slope and y-axis perpendicular to it.

For ΣF_x:

F_f = μ ⋅ N ... friction force
F ... girl's pulling force → why do you multiply it with the sine of the hill's angle?
plus: you forgot, that the weight has a component in x-direction to

ΣF_y looks good to me
The length of the slope L is calculated correctly.

When calculating the work why do you multiply it with the cosine of θ? The girl is pulling the sleigh parallel to the slope, you already calculated the length L, which you need for the calculation of the work.

1) Think about the sum of the forces in x-direction again:
a) direction of the girl's pulling force (or at least the component that is able to do work)
b) the component of the weight in x-direction

2) To calculate the work multiply the pulling force with the length the sleigh is actually pulled (the whole slope)
That is correct and it must be less than the work of the girl, else you create a perpetuum mobile.

OK, doing it your way i get:
In the steep way
F=46N
W= 9200 J

The not steep way:
F=27.86 N
W=10755.61 J

 

[haruspex]

OK, doing it your way i get:
In the steep way
F=46N
W= 9200 J

The not steep way:
F=27.86 N
W=10755.61 J

Looks right.

 

[stockzahn]

OK, doing it your way i get:
In the steep way
F=46N
W= 9200 J

The not steep way:
F=27.86 N
W=10755.61 J

That's it, well done!

 

[vetsi]

You are right, i used m=8 on my calculator instead of m=38.

In the task it says "How much work does the child have to do on the sleigh". Shouldn't you then use 8kg as the mass?

 

[stockzahn]

In the task it says "How much work does the child have to do on the sleigh". Shouldn't you then use 8kg as the mass?

It was the calculation of the friction force in the 2nd part: When the girl used the sleigh to slide down. As she is sitting on the sleigh getting downwards the mass is the sum of the masses of the girl and the sleigh, hence 38 kg.

 

[haruspex]

It was the calculation of the friction force in the 2nd part: When the girl used the sleigh to slide down. As she is sitting on the sleigh getting downwards the mass is the sum of the masses of the girl and the sleigh, hence 38 kg.

... and anyway, the mass used would be irrelevant in b).

 

[vetsi]

okey. I thought that you used it in a

 

[physicsbrained]

in a): I thought the work done would be the same either jo drag the sleigh the steep- or not-so-steep way, cause the height is the same.

 

[physicsbrained]

in a): I thought the work done would be the same either jo drag the sleigh the steep- or not-so-steep way, cause the height is the same.

Missed the friction, sorry