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A stone of mass m = 1.05 kg is released from a height of h = 2.1 m into a pool of water. At a time of t = 1.83 s after hitting the surface of the water, the stone's velocity has decreased by 50%.
What is the magnitude of the average force the stone experiences, in newtons, during the time t?

Respuesta :

Answer:

Explanation:

ignoring air resistance, the kinetic energy at water impact will equal the potential energy converted

½mv² = mgh

v = √(2gh)

v = √(2(9.81)2.1) = 6.4188... m/s

after impact, an impulse will result in a change of momentum.

There is a downward impulse due to gravity equal to the weight of the stone and an upward average force due to water resistance and buoyancy force.

FΔt = mΔv

(F - mg)Δt = m(vf - vi)

(F - mg) = m(vf - vi)/Δt

F = m(vf - vi)/Δt + mg

F = m((vf - vi)/Δt + g)

F = 1.05(((½(-6.4188) - -6.4188)/ 1.83) + 9.81)

F = 12.14198...

F = 12.1 N

The magnitude of the average force the stone experiences, will be 12.1 N.

What is force?

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The given data in the problem is;

m is the stone mass = 1.05 kg

The height is h, = 2.1 m i

The time is, t = 1.83 s

Velocity decreased by 50%.

From the law of conservation of energy the kinetic energy is equal to the potential energy;

[tex]\rm \frac{1}{2} mv^2 = mgh \\\\ v = \sqrt{2gh}\\\\ v = \sqrt{2 \times 9.81 \times 2.1 } \\\\ v= 64188 \ m/sec.[/tex]

From Newton's second law of motion, the impulsive force is equal to the change in momentum;

[tex]\rm F \triangle T = m \traingle v \\\\ (F - mg) = \frac{v_f-v_i}{\triangle t } \\\\ F = \frac{v_f-v_i}{\triangle t } + mg \\\\ F =m\frac{v_f-v_i}{\triangle t + g} \\\\ F= 12.14 98 \\\\ F=12.1 N[/tex]

Hence the magnitude of the average force the stone experiences, will be 12.1 N.

To learn more about the force refer to the link;

https://brainly.com/question/26115859