# Classic box up a hill problem

1. Aug 17, 2011

### Ciricus

1. The problem statement, all variables and given/known data
A 10kg Box is being pushed up a 3m long hill at the angle of 30 degrees. A constant force is being applied to the box parallel to the ramp. The box starts from rest. For questions a and b we can assume no friction.

a) If you were to get the box to the top of the hill in 5.0s, what acceleration must it have?

b) What is the nett force on the box if it is accelerating at this rate?

c) If the coefficient of kinetic friction between the box and the hill is 0.42, what force is required to push the box?

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2. Relevant equations (or so I believe)

a) d = Vit + $\frac{1}{2}$at2

b) Fx = mgsin($\Theta$)
Fy= mgcos($\Theta$)

c) f =$\mu$N

3. The attempt at a solution

a) d = Vit + $\frac{1}{2}$at2
Since Vi = 0m/s d = $\frac{1}{2}$at2

To solve for a: 2 $\times$ $\left(\frac{d}{t^2}\right)$
a = 2 $\times$ $\left(\frac{3m}{5s^2}\right)$
a = 0.24m/s2

b) x forces: Fx = mgsin($\Theta$)
Fx = mgsin(30)
Fx = 10kg $\times$ 0.24m/s2 $\times$ sin(30)
Fx = 1.2N

y forces: Fy = mgcos($\Theta$)
Fy = mgcos(30)
Fy = 10kg $\times$ 9.8m/s2 $\times$ cos(30)
Fy = 84.87N

$\therefore$ Nett force = Fx + Fy = 86.07N

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C) $\mu$ = $\frac{f}{n}$
0.42 = $\frac{f}{1.2}$
Which gives 0.5N

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I lost faith in my ability to do it after b). I'm not sure if you're supposed to be using 9.8m/s2 or 0.24m/s2 when finding the value for mgsin($\Theta$). If i use 9.8m/s2 i get a more reasonable answer of 42N, however i'm not applying the 'new acceleration' into the equation at all.

Thanks for any help/direction with this in advance!

2. Aug 18, 2011

### alphali

i think u should solve b as follows :
$\sum$F=ma(here u need to use the acceleration)
F+W=ma
(F is the force u need to find and W is the force of gravity or weight)
projection along x-axis
f-Wx=ma ( -Wx because the force is acting in the opposite direction)
Wx=g.m.sin30=9.8$\times$10$\times$sin30=49N
THEN
F=Wx+ma=49+(10)(0.24)=51.4N

Last edited: Aug 18, 2011