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A ball whose mass is 0.3 kg hits the floor with a speed of 5 m/s and rebounds upward with a speed of 2 m/s. If the ball was in contact with the floor for 1.5 ms (1.5multiply10-3 s), what was the average magnitude of the force exerted on the ball by the floor?

Respuesta :

opudodennis

Answer:

1400 N

Explanation:

Change in momentum equals impulse which is a product of force and time

Change in momentum is given by m(v-u)

Equating this to impulse formula then

m(v-u)=Ft

Making F the subject of the formula then

[tex]F=\frac {m(v-u)}{t}[/tex]

Take upward direction as positive then downwards is negative

Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then

[tex]F=\frac {0.3(2--5)}{0.0015}=1400N[/tex]

asuwafohjames

The average force exerted on the ball by the floor is  1400 N.

What is force?

Force is the product of mass and acceleration. The S.I unit of force is Newton (N)

To calculate the average magnitude of the force exerted on the ball by the floor, we use the formula below.

Formula:

  • F = m(v-u)/t............... Equation 1

Where:

  • F = Force exerted on the ball by the floor.
  • m = mass of the floor
  • v = final velocity of the ball
  • u = initial velocity of the ball
  • t = time

From the question,

Given:

  • m = 0.3 kg
  • v = 5 m/s
  • u = -2 m/s (rebounds)
  • t = 1.5×10⁻³ s

Substitute these values into equation 1

  • F = 0.3[(5)-(-2)]/(1.5×10⁻³)
  • F = 0.3(5+2)/(1.5×10⁻³)
  • F = 2.1/(1.5×10⁻³)
  • F = 1.4×10³
  • F = 1400 N

Hence, the average force exerted on the ball by the floor is  1400 N.

Learn more about force here: https://brainly.com/question/13164598

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