A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force of friction on the cart
1 answer:
Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.
We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force
acting in the opposite direction. So Newton's second law can be rewritten as
or
since the frictional force is 15 N and we want to achieve an acceleration of
, we can substitute these values to find what is the force the man needs:
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.
We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force
or
since the frictional force is 15 N and we want to achieve an acceleration of
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Answer:
The force per unit length (N/m) on the top wire is 16.842 N/m
Explanation:
Given;
distance between the two parallel wire, d = 38 cm = 0.38 m
current in the first wire, I₁ = 4.0 kA
current in the second wire, I₂ = 8.0 kA
Force per unit length, between two parallel wires is given as;
where;
μ₀ is constant = 4π x 10⁻⁷ T.m/A
Substitute the given values in the above equation and calculate the force per unit length
Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m