The Mysteries of (a) and (c): Homework Solutions

[goluigi2196]

1. Homework Statement

A box of books with a mass of 10.0 kg is shoved across the floor by a force of 415 N exerted downward at an angle of 35° above the horizontal.

(a) If µk between the box and the floor is 0.23, find the normal force on the box.
For (a), I got 247.834 N

(b) Find the acceleration of the box

(c) How long does it take to move the box 4 m, starting from rest?

I'm stuck on (b) and (c)

Fp is the pushing force. Fpx is the x direction and Fpy is the y direction.
ax is the acceleration in the x direction

2. Homework Equations :
sigmaFx=max
sigmaFy=0

solving for ax, I get:
ax=(-Ff-Fpx)/m3. The Attempt at a Solution :

ax=(-247.834-415cos35)/10
ax=-38.876 m/s^2

I'm not really sure if I'm supposed to get a negative number ?

For (c), I am absolutely stuck.

 

[jdismu1]

b)
First youre forgetting an important equation, force of friction
F_f=\mu_k*F_n = 0.23 * 247.834 N = 57.002 N
you used the normal force instead of the force of friction

next find the force of the push F_p

because it is downward at a 35^o angle only a portion of the force is applied in the x direction.
F_px = 415cos(35^o) = 339.95

Set up a free body diagram for the problem

F_px<----(box)-->F_f
set the xy plane on the box, with the box centered at the origin, and youll see that F_p is in the negative direction and F_f is in the positive, when plugging this in the net force equation make sure to include signs

You want to find the acceleration, so you have to find the net force on the box and set that equal to ma_x

F_net=\SigmaF
\SigmaF=-F_p+F_f

next F_net= -339.95 N +57.002 N = -282.95 N is youre net force

use this to find a_x

F_net/m = a_x
-282.95 N/ 10g = -28.295 m/s²

the negative is simply for direction, if they ask for magnitude give them 28.295 however if they want direction the negative is necessary

 

[jdismu1]

c)
once you have acceleration you have to use one of the motion equations
what you need is time and you have v₀ (initially velocity), acceleration, and distance

use the equation

d=v₀t+(1/2)at²

solve for t
luckily we know that the box starts at rest so v₀ = 0 and that portion of the equation disappears

so now we have
d=(1/2)at²

solve for t and youll find

square root(2d/a)=t <-- sorry something is wrong with the forums can't do the square root deal

finally plug and chug

square root(2*4 m/ 28.295 m/s^2) = 0.5317 s <-- again something wrong with the forums XD

what about the negative acceleration? technically the box is moving in the direction of the force therefore the displacement is -4 m and the acceleration is also negative. negative divided by negative is positive and it works.

hope this helps!
first time answering a question on here

 

[goluigi2196]

Hey thanks man. That helped a lot. I usually get physics but friction is something that I have a hard time with.

 

[jdismu1]

anytime