A man does 500 J of work pushing a car a distance of 2 m. How much force does he apply? Assume there is no friction.
2 answers:
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Answer:
a) m=20000Kg
b) v=0.214m/s
Explanation:
We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.
For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as
, and the mass of the first and second coals as
and
respectively
We start with the transition between parts A and B, so we have:
Which means
And since we want the mass of the first coal thrown () we do:
Substituting values we obtain
For the transition between parts B and C, we can write:
Which means
Since we want the new final speed of the car () we do:
Substituting values we obtain
Answer:
Explanation:
From the Question we are told that:
Mass
Coefficient of kinetic friction
Generally the equation for Frictional force is mathematically given by
Generally the Newton's equation for Acceleration due to Friction force is mathematically given by
Therefore