First, we need to calculate the force of friction acting on the computer as it is dropped on the conveyor belt. We can do this using the equation:
Ffriction = uk * Fnormal
where Fnormal is the normal force acting on the computer (equal to its weight, mg) and uk is the coefficient of kinetic friction.
Fnormal = mg = 10 kg * 9.81 m/s^2 = 98.1 N
Ffriction = 0.30 * 98.1 N = 29.4 N
Next, we can use the equation for constant acceleration to determine how far the computer is dragged before it reaches a constant speed on the conveyor belt:
d = (v^2 - u^2) / (2a)
where d is the distance traveled, v is the final velocity (equal to the speed of the conveyor belt, 3.1 m/s), u is the initial velocity (0 m/s), and a is the acceleration.
We can calculate the acceleration using the equation:
Fnet = ma
where Fnet is the net force acting on the computer (equal to the force of friction), and m is the mass.
Fnet = Ffriction = ma
a = Fnet / m = 29.4 N / 10 kg = 2.94 m/s^2
Now we can plug in the values:
d = (3.1^2 - 0^2) / (2 * 2.94) = 1.69 m
Therefore, the computer is dragged 1.69 meters before it reaches a constant speed on the conveyor belt.
A factory robot drops a 10kg computer onto a conveyer belt running at 3.1m.s. the material are such as us=0.5 and uk=0.30 between the belt and the computer.how far is the computer dragged before it is ridding smoothly os the belt
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1 answer