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

The force the swimmer applied when he pushes on the pool wall is 490 N.

The given parameters;

The force applied by the swimmer is calculated by applying Newton's second law of motion as shown below;

F = ma

∑F = 0

F - Fk = ma

F = Fk + ma

F = 100 + (65 x 6)

F = 490 N

Thus, the force the swimmer applied when he pushes on the pool wall is 490 N.

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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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