# An accelerating conveyor drags a box

1. Aug 21, 2015

### Karol

1. The problem statement, all variables and given/known data
The velocity of a conveyor is V=30+0.1t2. a box of mass m=3[kg] falls from zero height at t=0. the kinetic coefficient of friction is μk=0.4 and the static is μs=0.5.
1) At which times the friction is kinetic and at which static
2) What's the box's velocity in the different times
3) What's the distance the box has traveled relative to the conveyor from t=0 to static friction
4) What's the work that the conveyor has done from t=0 to static friction
5) What's the work the box has done on the conveyor from t=0 to static friction
6) What's the work the conveyor has done in the static friction phase

2. Relevant equations
Distance: $x=\int vdt$
Distance at constant acceleration: x=at2

3. The attempt at a solution
The velocities of the conveyor and box must equal in order that the friction will be static. v is the box's velocity:
$$v=at=g\mu_k\cdot t$$
$$30+0.1t^2=g\mu_k\cdot t\Rightarrow t_1=20,\; t_2=30$$
I choose t=20. the maximum acceleration the box can stand in the static friction is $a=g\mu_s=0.5\cdot 10=5$
When the conveyor's acceleration exceeds that value the friction will become kinetic again. the conveyor's acceleration:
$$V'=(30+0.1t^2)'=0.2t,\; 0.2t=5\Rightarrow t=25$$
Between t=20 and t=25 the friction will be static. the box's velocity in the first kinetic stage is $v=at=g\mu_k t$. in the static stage the conveyor's velocity is $v=V=30+(t-20)^2$.
In the second kinetic stage it's the conveyor's velocity minus the box's velocity relative to the conveyor. the box's velocity i find from a coordinate system attached to the conveyor. d'allamber's force is the conveyor's acceleration, and the total force is d'alambers's minus friction: $F=m 0.2t-mg\mu_k$. the speed is the integral of acceleration, so:
$$v=30+0.1t^2-\int_{25}^t 0.2t-g\mu_k dt=4t-7.5$$
v' is the box's velocity relative to the conv': $v=V+v'$, the distance the box has passed, relative to the conv', from t=0 to the static stage is the integral of the velocity:
$$v'=v-V=g\mu_kt-(30+0.1t^2)=4t-30-0.1t^2,\; x=\int_0^{20} 4t-30-0.1t^2 dt=66\frac{2}{3}$$
It should be negative. The work the conveyor has done from t=0 to static is FxS, S is the distance the box traveled:
$$W=mg\mu_k\cdot \frac{1}{2}g\mu_k t^2=96,000[Joule]$$
The work the box has done on the conv' during that time is the same force times the distance the conv' traveled. the distance it traveled is the integral of the velocity:
$$W=mg\mu_k\cdot\left( 30t+\frac{1}{30}t^3 \right)=10,400[Joule]$$
Is this work the loss of energy? is it all transformed into heat?
At the static friction stage the conv' has done work:
$$W=mg\mu_s\cdot \left[ 30(25-20)+\frac{1}{30}(25-20)^3 \right]=6062.5[Joule]$$

Last edited: Aug 21, 2015
2. Aug 21, 2015

### Nathanael

Equation is good, but t=20 isn't a solution.

Why do you shift the time-zero to 20 seconds? The unshifted equation already describes the speed of the conveyor at all t.

3. Aug 21, 2015

### Karol

$$30+0.1t^2=g\mu_k\cdot t\Rightarrow t_1=10,\; t_2=30$$
$$v=V=30+t^2$$

4. Aug 22, 2015

### Nathanael

I didn't notice at first but you're also missing the 0.1 factor.
But yeah, the speed of the box during the static-friction-stage would just be the equation given in the problem statement.

The force during the static friction stage would not be always be $mg\mu_s$

5. Aug 22, 2015

### Karol

$$v=V=30+0.1t^2$$
the distance the box has passed, relative to the conv', from t=0 to the static stage is the integral of the velocity:
$$v'=v-V=g\mu_kt-(30+0.1t^2)=4t-30-0.1t^2,\; x=\int_0^{10} 4t-30-0.1t^2 dt=-133\frac{1}{3}$$
At the static friction stage the conv' has done work:
$$W=F\cdot S=ma\cdot x=m\cdot 0.2t\left( 30t+\frac{1}{30}t^3 \right)=\int_{10}^{25} 18t^2+0.02t^4=126,412.5[Joule]$$

6. Aug 22, 2015

### Nathanael

The true equation would be $W=\int F\cdot dS=$ ... only in the case of constant F does this simplify to what you wrote $\Big (\int F\cdot dS=F\cdot \int dS=F\cdot S \Big )$.

But in our situation the force is not constant.

7. Aug 22, 2015

### Karol

Thanks, so i understand the rest is good?
I made a mistake in the works done by the conveyor and the box. the work done by the conv':
$$W=mg\mu_k\cdot \frac{1}{2}g\mu_k t^2=240,000[Joule]$$
The work done on the conv':
$$W=mg\mu_k\cdot\left( 30t+\frac{1}{30}t^3 \right)=400,000[Joule]$$
And i understand that the difference is the energy loss, right?

8. Aug 22, 2015

### Nathanael

Yep everything else looks good. (Haven't checked the numbers.) Your understanding is right.

The only mistake I see is in your calculation of the work done during the static-friction-stage.

9. Aug 23, 2015

### Karol

$$W=F\cdot S=ma\cdot x=m\cdot 0.2t\left( 30t+\frac{1}{30}t^3 \right)=\int_{10}^{25} 18t^2+0.02t^4=474,375[Joule]$$
Thanks Nathanel

10. Aug 23, 2015

### Nathanael

What you're doing here doesn't make sense to me. (What are you integrating? There's no differential, it doesn't make sense.)

You can't use $W=F\cdot S$ because F varies in time. Use $W=\int F\cdot dS = \int F\cdot Vdt$

11. Aug 23, 2015

### Karol

$$W=\int F\cdot vdt=\int_{10}^{25} 0.2mt\left( 30+0.1t^2 \right)dt=\int_{10}^{25} 18t+0.06t^3dt=10,434.4[Joule]$$

12. Aug 23, 2015

### Nathanael

Ok, looks good now.

13. Aug 23, 2015

### Karol

Thanks Nathanel