A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force of 77.0 N is required to set the block in motio
n, after which a horizontal force of 56.0 N is required to keep the block moving with constant speed.Find the coefficients of static and kinetic friction from this information.
1 answer:
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The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.
We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force
acting against the motion. Since their resultant must be zero, we have:
The frictional force is
where
is the coefficient of kinetic friction
is the weight of the block.
Substituting these values, we find the magnitude of the force F:
We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force
The frictional force is
where
Substituting these values, we find the magnitude of the force F:
Answer:
Explanation:
For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:
where the suffix i means initial, and the suffix f means final.
The initial momentum will be:
as the second puck is initially at rest:
Using the unit vector pointing in the original line of motion:
So:
Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula
So, our velocity vectors will be:
We got
So, we got the equations:
and
.
From the last one, we get:
and, for the first one:
so:
and