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STALIN [3.7K]
2 years ago
12

Calculate the force required to accelerate a 600 g ball from rest to 14 m/s in 0.1 s.

Physics
1 answer:
Evgen [1.6K]2 years ago
5 0

<u>Statement</u><u>:</u>

A force is required to accelerate a 600 g ball from rest to 14 m/s in 0.1 s.

<u>To </u><u>find </u><u>out</u><u>:</u>

The force required to accelerate the ball.

<u>Solution</u><u>:</u>

  • Mass of the ball (m) = 600 g = 0.6 Kg
  • Initial velocity (u) = 0 m/s [it was at rest]
  • Final velocity (v) = 14 m/s
  • Time (t) = 0.1 s

  • Let the acceleration be a.
  • We know the equation of motion,
  • v = u + at

  • Therefore, putting the values in the above formula, we get
  • 14 m/s = 0 m/s + a × 0.1 s
  • or, 14 m/s ÷ 0.1 s = a
  • or, a = 140 m/s²

  • Let the force be F.
  • We know, the formula : F = ma

  • Putting the values in the above formula, we get
  • F = 0.6 Kg × 140 m/s²
  • or, F = 84 N

<u>Answer</u><u>:</u>

The force required to accelerate the ball is 84 N and this force acts along the direction of motion.

Hope you could understand.

If you have any query, feel free to ask.

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A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
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The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

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<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

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