Calculating force needed to push sled down incline

AI Thread Summary
To calculate the force needed to push a 22kg sled down a 6-degree incline at 60 km/hr after 75m, the net force equation Fnet = ma is applied alongside kinematic equations. The sled's acceleration is determined to be 1.85 m/s² by converting the final speed to m/s and using the formula (V2)² = (V1)² + 2a(d2-d1). The frictional force is calculated using Ff = uN, where the coefficient of friction is 0.20. A common error noted is the handling of signs in force equations, emphasizing the importance of consistent sign conventions. Correctly applying these principles leads to the accurate calculation of the applied force required.
jmm5180
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Homework Statement


Calculate the force required to push a 22kg sled down a 6 degree hill and have an ending speed of 60km/hr after traveling 75m when the coefficient of friction is 0.20.

Homework Equations


Fnet= ma
(V2)^2 = (V1)^2 + 2a(d2-d1)
Ff= uN
W= mg

The Attempt at a Solution


I'll use @ for theta, and u for mu. Fa= force applied to the sled, Ff=force of friction

With N lying on the positive Y axis and Weight in the x direction lying on the positive x axis, I set Wy=wcos@ and Wx= wsin@, so N=wcos@.

I tried summation of forces:
xFnet = Wx + Fa - Ff = ma
Fa= Ff - Wx + ma
Fa = uN - wsin@ + ma
Fa = u(wcos@) - wsin@ + ma
Fa = umgcos@ - mgsin@ + ma
Fa = m(ugcos@ - gsin@ + a)

I then tried to solve for a using (V2)^2 = (V1)^2 + 2a(d2-d1)
Since the starting velocity was 0 and the starting distance is 0,
A = (V2)^2 / (2d2)

I converted 60 km/hr to m/s, to get 16.66 m/s.
A= (16.66^2)/ 2(75) = 1.85 m/s^2

Plugging that in...

Fa= 22 ( (0.2)(-9.8)(cos6) - (-9.8sin6) + 1.85 ))
I got Fa = 20.36 N, though that isn't correct. I'm not sure what I did wrong.
 
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jmm5180 said:
Fa= 22 ( (0.2)(-9.8)(cos6) - (-9.8sin6) + 1.85 ))
Careful with signs. g is just a positive constant.
 
Check the signs of the forces. Looks like you have the required net force (ma) the same sign as the friction force (uN)?

May not be the only problem but that looks wrong to me.
 
Doc Al said:
Careful with signs. g is just a positive constant.

Thank you. I've used -9.8 m/s^2 before in questions and have gotten the correct answer. How do I know when to use the positive/negative version?
 
jmm5180 said:
Thank you. I've used -9.8 m/s^2 before in questions and have gotten the correct answer. How do I know when to use the positive/negative version?
I would always use g as a positive constant and add signs as needed. For example, using down as negative, the acceleration of a falling object is -g = -9.8m/s^2.

When you set up your force equations, always assign your signs consistent with whatever sign convention you are using. (For example, taking down the incline as positive would make the weight component, the applied force, and the acceleration positive and friction negative.)
 
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