How Do You Calculate Kinetic Friction and Motion Distance in a Physics Problem?

[oMovements]

Homework Statement

An electric motor is used to pull a 125kg box across a floor using a long cable. The tension in the cable is 350N and the box accelerates at 1.2 m/s/s [fwd] for 5.0s. The cable breaks and the box slows down and stops
a )Calculate the coefficient of kinetic friction. [ans:0.16]
b) How far does the box travel up to the moment the cable breaks? [ans: 15m]
c) How far does the box travel from the moment the cable breaks until it stops? [ans: 11m]

Homework Equations

Kinematics equations

The Attempt at a Solution

F_t-F_k=ma
350-F_k=(125)(1.2)
F_k=200N

μ_k=F_k/F_n
=200/(125)(9.8)
μ_k=0.16

b) D=V_ft-1/2at^2
= 0-1/2(-1.2)(5)^2
= 15m

c)

I know my solution for b) is correct. But I don't understand why the V_f is zero when the cable breaks because shouldn't the V_f be zero when the box actually comes to a stop and not when the cable breaks. Assuming I solved b) correct, and Vf = 0 then how would I solve c) when the box actually comes to a stop?

 

[Doc Al]

b) D=V_ft-1/2at^2

Your formula is a bit off. That should be: D = v_i + 1/2at², where v_i is the initial velocity. (But your answer is correct, just the same.)

You'll need a different formula to calculate the distance the box slides after the cable breaks. (Or combine a few formulas.)

 

[oMovements]

Your formula is a bit off. That should be: D = v_i + 1/2at², where v_i is the initial velocity. (But your answer is correct, just the same.)

You'll need a different formula to calculate the distance the box slides after the cable breaks. (Or combine a few formulas.)

I don't understand why the initial velocity is zero when the cable breaks. Also what formula would I use?

 

[arildno]

In b), you are considering the time interval from t=0, to t=5.
It is at t=5 the cable breaks, but t=0 determines the initial velocity to be 0

 

[arildno]

You may, of course, use 0<=t<=5:
$$v_{f}=v_{0}+at\to{v_{0}}=v_{f}-at$$
Thus, you can rewrite the standard formula for distance as:
$$d=v_{f}t-\frac{1}{2}at^{2}$$
rather than:
$$d=v_{0}t+\frac{1}{2}at^{2}$$

a>0
--
In your solution at b), you made two errors that canceled each other's effects:
You set v_f=0, and a=-1.2
Thus, you got the correct result, but in an incorrect manner.

 

[oMovements]

You may, of course, use 0<=t<=5:
$$v_{f}=v_{0}+at\to{v_{0}}=v_{f}-at$$
Thus, you can rewrite the standard formula for distance as:
$$d=v_{f}t-\frac{1}{2}at^{2}$$
rather than:
$$d=v_{0}t+\frac{1}{2}at^{2}$$

a>0
--
In your solution at b), you made two errors that canceled each other's effects:
You set v_f=0, and a=-1.2
Thus, you got the correct result, but in an incorrect manner.

Thanks, I understand that. But how would I solve c), would time still be 5s when it comes to a stop?

 

[arildno]

Thanks, I understand that. But how would I solve c), would time still be 5s when it comes to a stop?

No, it would not.
You do not strictly need to find it, though, since for the new distance, you have the new initial&final velocities, plus the new acceleration.

 

[oMovements]

No, it would not.
You do not strictly need to find it, though, since for the new distance, you have the new initial&final velocities, plus the new acceleration.

How would I find these new values?

 

[Doc Al]

I don't understand why the initial velocity is zero when the cable breaks.

It's not. The box is moving when the cable breaks. (The final velocity will be zero.)

Also what formula would I use?

Here's a list of standard kinematic formulas: Basic Equations of 1-D Kinematics

 

[arildno]

How would I find these new values?

You might ask yourself first:
1. What forces act upon the car just after the cable breaks and beyond?
2. What is the velocity of the car just after the cable breaks (or in that same moment if you like!), and how can you calculate it?
3. What is the velocity of the car when it finally stops?

 

[oMovements]

You might ask yourself first:
1. What forces act upon the car just after the cable breaks and beyond?
2. What is the velocity of the car just after the cable breaks (or in that same moment if you like!), and how can you calculate it?
3. What is the velocity of the car when it finally stops?

Since the cable breaks, the net force would become F_k=200N since there is no tension.
a= Fnet/m
= 200/125
a= 1.6 m/s/s

To find the velocity when the cable breaks, I would have to determine the final velocity when t=5 and acceleration is 1.2m/s/s

a = vf-vi/t
1.2 = Vf-0/5
Vf = 6 m/s

This means this would now become the initial velocity after the cable breaks.

So now I have enough information to determine the distance it travels after the cable breaks and comes to a stop.

Vf²=Vi²+2ad
d = Vf²-Vi₂ / 2a
= 0-(6)²/2(-1.6)
= -36/-3.2
= 11.25
= 11m
So is this correct? The answer that's given is rounded to 2 significant digits

 

[arildno]

yup!