What is the required force to push a 5.00-kg box up a 30° incline in 4.00s?

[Zarrey]

Homework Statement

A 5.00-kg box sits at rest at the bottom of a ramp that is 8.00m long and that is inclined at 30.0° above the horizontal. The coefficient of kinetic friction is μ_k = 0.40, and the coefficient of static friction is μ_s = 0.50. What constant force F, applied parallel to the surface of the ramp, is required to push the box to the top of the ramp in a time of 4.00s?g = 9.8 in my class

Homework Equations

F_g = m(g)
F_N = m(g)(cos(θ))
F_fk = μ_k(F_N)
F_NET = m(a)

The Attempt at a Solution

F_g = 5(9.8) = 49N

F_N = 5(9.8)(cos(30)) = 42.44N

F_fk = .40(42.44) = 16.98N

This is where I get stuck. I am able to find the minimum amount required to get it to start and keep moving but not what the question is asking. My textbook does not explain this topic very well so I was getting frustrated.

 

[Infinitum]

Hi Zarrey!

You know the forces that act on the block along(parallel) the incline. These are constant, and so is your force F. Now the sum of these forces need to produce an acceleration a, such that the distance traveled in t seconds is the length L of the incline. As an equation, you can write that as

$$L = \frac{1}{2}at^2$$

 

[Zarrey]

so then

L = .5a(t²)
8 = .5a(4²)
8 = .5a(16)
8 = 8a
1m/s = a

F_ll = 49(sin(30))
F_ll = 24.5N

F_NET = F_p - F_ll - F_fk

And then Newtons law:

F_p - F_ll - F_fk = m(a)
F_p - 24.5 - 21.22 = 5(1)
F_p = 50.72N

is this correct? (i substituted F_p as the push force since that what we use in class)

 

[Zarrey]

Bump.

Sorry for this bump but this question is kind of important as it is on a take home test. I'm kind of freaking out because I am already turning it in late. A validation response would be much appreciated. And thanks for the help in advance.

 

[Doc Al]

so then

L = .5a(t²)
8 = .5a(4²)
8 = .5a(16)
8 = 8a
1m/s = a

F_ll = 49(sin(30))
F_ll = 24.5N

F_NET = F_p - F_ll - F_fk

And then Newtons law:

F_p - F_ll - F_fk = m(a)
F_p - 24.5 - 21.22 = 5(1)
F_p = 50.72N

is this correct? (i substituted F_p as the push force since that what we use in class)

Looks good except that you used static friction instead of kinetic friction.

 

[Infinitum]

so then

L = .5a(t²)
8 = .5a(4²)
8 = .5a(16)
8 = 8a
1m/s = a

F_ll = 49(sin(30))
F_ll = 24.5N

F_NET = F_p - F_ll - F_fk

And then Newtons law:

F_p - F_ll - F_fk = m(a)
F_p - 24.5 - 21.22 = 5(1)
F_p = 50.72N

is this correct? (i substituted F_p as the push force since that what we use in class)

Hi, sorry for the late reply.

The bolded term seems wrong. You need to use the kinetic friction coefficient.

 

[Zarrey]

thanks a lot, i appreciate everyone's help in getting me to understand this frustrating topic :).