1. physicis problem #2

1) A box of books weighing 200 N is shoved across the floor of an apartment by a force of 300 N exerted horixontally.
if the cofficient of kinetic friction between the floor anf box is 0.52, how long does it take the box th move 3.0m starting fron rest

I want the concepts to solve this queation

2. Originally Posted by r-soy
1) A box of books weighing 200 N is shoved across the floor of an apartment by a force of 300 N exerted horixontally.
if the cofficient of kinetic friction between the floor anf box is 0.52, how long does it take the box th move 3.0m starting fron rest

I want the concepts to solve this queation
Let $\displaystyle F$ = applied force of 300 N

$\displaystyle f$ = force of kinetic friction

$\displaystyle N$ = Normal force acting on the box

$\displaystyle W$ = weight of the box (200 N)

(note that $\displaystyle N = W$ in this case)

$\displaystyle \mu$ = coefficient of friction (0.52)

$\displaystyle m$ = mass of the box (not the weight)

$\displaystyle a$ = acceleration of the box

$\displaystyle F_{net} = F-f$

$\displaystyle ma = F - \mu \cdot N$

$\displaystyle ma = F - \mu \cdot W$

$\displaystyle \displaystyle a = \frac{F - \mu \cdot W}{m}$

once you've calculated the acceleration ...

$\displaystyle \Delta x = \frac{1}{2} at^2$ ... solve for $\displaystyle t$

3. a = 300-(0.52)(200)/400
= 299

a=1/2(299)t^2
t = 0.04

Is ok ?

4. No. Could you show us how you got the mass of the box to be 400 kg?

Later, you should get the time as 2.5 s.

a=1/2(299)t^2
t = 0.04

Is not good either.

$\displaystyle \Delta x = \dfrac12 at^2$

$\displaystyle 30 = \dfrac12 (299)t^2$

$\displaystyle t^2 = \dfrac{30 \times 2}{299} = 0.2$

$\displaystyle t = \sqrt{0.2}= 0.45 s$

0.45 ??

6. No, the 0.45 s was the reply concerning your wrong calculation so that you know where you did a wrong calculation so that next time, you know what to do using the correct values.

I told you to show us how you got 400 kg for the mass.

7. How i can find the mass ??

i find mass 2 X 200 = 400

8. And can I know why you did that?

Let me remind you that:

$\displaystyle W = mg$

W is the weight, m is the mass and g the acceleration due to gravity.

9. a = 300-(0.52)(200)/20
=9.8

t^2 = 30 X 2 /9.8 = 6.12
t = 2.4

now is ok ??

10. Right, but you usually have to round it off well. All the numbers are 2.4743583 which gives 2.47 or 2.5 seconds.

I thought you would have taken g as 9.8 m/s^2... but nevermind. If you are told to use g = 10m/s^2, you use it as such.