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BigorU [14]
3 years ago
13

A 65-kg swimmer pushes on the pool wall and accelerates at 6 m/s^2. The friction experienced by the swimmer is 100 N. How many N

of force does the swimmer applies a force of
Physics
1 answer:
Helga [31]3 years ago
5 0

Answer:

490N

Explanation:

According Newton's second law!

\sum Force = mass × acceleration

Fm - Ff = ma

Fm is the moving force

Ff s the frictional force = 100N

mass = 65kg

acceleration = 6m/s²

Required

Moving force Fm

Substitute the given force into thr expression and get Fm

Fm -100 = 65(6)

Fm -100 = 390

Fm = 390+100

Fm = 490N

Hence the force that will cause two cart to move is 490N

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2 years ago
Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot
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Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0 (Ec. 1)

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Where:

P - Pushing force, measured in newtons.

T - Tension, measured in newtons.

\mu_{k} - Coefficient of kinetic friction, dimensionless.

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W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P+T -\mu_{k}\cdot W = 0

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\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

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