- #1
Karol
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Homework Statement
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
Homework Equations
Distance: ##x=\int vdt##
Distance at constant acceleration: x=at2
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:
$$W=mg\mu_s\cdot \left[ 30(25-20)+\frac{1}{30}(25-20)^3 \right]=6062.5[Joule]$$
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