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A 51.5-kg swimmer with an initial speed of 1.25 m/s decides to coast until she comes to rest. If she slows with constant acceleration and stops after coasting 2.20 m, what was the force exerted on her by the water?

Respuesta :

xero099

Answer:

[tex]F_{water} = -18.282\,N[/tex]

Explanation:

The acceleration of the swimmer is obtained by this expression:

[tex]a = \frac{v_{f}^{2} - v_{o}^{2}}{2\cdot s}[/tex]

[tex]a = \frac{(0\,\frac{m}{s})^{2} - (1.25\,\frac{m}{s} )^{2}}{2\cdot (2.20\,m)}[/tex]

[tex]a = -0.355\,\frac{m}{s^{2}}[/tex]

The force exerted on her by the water is:

[tex]F_{water} = m\cdot a[/tex]

[tex]F_{water} = (51.5\,kg)\cdot (-0.355\,\frac{m}{s^{2}} )[/tex]

[tex]F_{water} = -18.282\,N[/tex]

asuwafoh

Answer:

-18.283 N.

Explanation:

Using the equation of motion,

v² = u²+ 2as................ Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance.

make a the subject of the equation

a = (v²-u²)/2s......... Equation 2

Given: u = 1.25 m/s, v = 0 m/s ( comes to rest), s = 2.20 m.

Substitute into equation 2

a = (0²-1.25²)/(2×2.2)

a = -1.5625/4.4

a = -0.355 m/s²

Note: The negative sign indicate that the swimmer is decelerating.

F = ma

Where F = force, m = mass of the swimmer.

Given: m = 51.5 kg.

F = 51.5(-0.355)

F = -18.283 N.

Note: The negative sign also mean that the force is in opposite direction to the motion of the swimmer

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