Answered

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A person pushes an object of mass 5.0 kg along the floor by applying a
force. If the object experiences a friction force of 10 N and accelerates at
18 m/s^2, what is the magnitude of the force exerted by the person?

Sagot :

elcharly64

Answer:

The magnitude of the force exerted by the person is 100 N

Explanation:

Net Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

Fn = ma

Where a is the acceleration of the object.

The net force is the sum of all forces exerted over a body. When an object is moved along a rough surface it experiences two horizontal forces and two vertical forces (provided there is no vertical component of the applied force).

The vertical forces are the Normal and the Weight and they are balanced, i.e.: N = W = mg.

The horizontal forces are The applied force (Fa) and the friction force (Fr). They are not balanced because the object is accelerated in that direction. The net force is:

Fn = Fa - Fr

Applying the first equation:

Fa - Fr = ma

Solving for Fa:

Fa = Fr + ma

Substituting the given values m=5 kg, Fr=10 N, [tex]a=18\ m/s^2[/tex].

Fa = 10 + 5*18 = 10 + 90 = 100

Fa = 100 N

The magnitude of the force exerted by the person is 100 N

baileyrd

Basically.

Given: F + m x a

Equation: 10 N + 5.0 kg x 18 m/s2

Solve: 100 N

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