Troubleshooting Pulling Force on a 20.0kg Sled

AI Thread Summary
A 20.0kg sled is pulled across a horizontal surface at a constant velocity with a pulling force of 80.0 N at a 30° angle. The initial calculation for the coefficient of kinetic friction was incorrect due to a mix-up between the vertical and horizontal components of the pulling force. The correct vertical component is 40 N, leading to a normal force of 156 N after adjusting for gravity. The accurate coefficient of kinetic friction is calculated as μ = 69.28 N / 156 N, resulting in approximately 0.444. The problem was resolved by correctly identifying the components of the pulling force.
l2udolph
Messages
22
Reaction score
0
A 20.0kg sled is being pulled across a horizontal surface at a constant velocity...

Hi. This is a problem on my physics homework that I can't seem to get the (correct) answer to. I would love to be told what I'm doing wrong. (:


This is a problem and what I did:

A 20.0kg sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 80.0 N and is directed at an angle of 30° above the horizontal. Determine the coefficient of kinetic friction.

m(sled) = 20kg
F(gravity) = 20kg * 9.8 m/s2 = 196 N

F(pull) = 80 N
F(pull,X) = 80sin30° = 40 N
F(pull,Y) = 80cos30° = 69.28 N

F(normal) = 196 N - 69.28 N = 126.72 N

Kinetic Friction = μ * F(normal)

μ = 40 N / 126.72 N = .3157

I looked in the back in the book, and apparently the correct answer is .444. What am I doing wrong?

Thanks!
 
Last edited:
Physics news on Phys.org


Nevermind, I figured out what I did incorrectly.
Lol, I mixed the vertical and horizontal components of the F(pull) by accident.
...

m(sled) = 20kg
F(gravity) = 20kg * 9.8 m/s2 = 196 N

F(pull) = 80 N
F(pull,Y) = 80sin30° = 40 N
F(pull,X) = 80cos30° = 69.28 N

F(normal) = 196 N - 40 N = 156 N

Kinetic Friction = μ * F(normal)

μ = 69.28 N / 156 N = .4441
 
Last edited:

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Pulling a sled across snow with a rope at an angle
Replies
8
Views
4K
2 Exercises with Forces -- dogs pulling a sled
Replies
29
Views
4K
Calculating Work and Change in Thermal Energy
Replies
2
Views
2K
Work and Kinetic Energy of a Sled
Replies
4
Views
5K
Force pulling on a string at an angle with a block at the other end
Replies
2
Views
1K
Pulling a 20.0kg Sled: Calculating the Coefficient of Friction
Replies
7
Views
7K
Finding work and the Change in Thermal Energy
Replies
5
Views
2K
Particle Newtonian Mechanics: Father Pulling 100 lb Sled
Replies
15
Views
2K
How Much Work Is Done When Pulling a Sled at an Angle?
Replies
3
Views
2K
Acceleration = (Force of boy on sled - Force of friction)/Mass of sled
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top