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4 years ago

A 60 kg bicyclist going 2 m/s increased his work output by 1,800 J. What was his final velocity?

Physics

2 answers:

Alinara [238K]4 years ago

6 0

<em>8 </em>was the correct answer, thanks

Bas_tet [7]4 years ago

3 0

Bicyclist initial kinetic energy is Ek=(1/2)*m*v² where m is his mass and v is his speed and that is equal to:

Ek=(1/2)*60*2²=120 J.

When we add the increased work output, we get the total kinetic energy:

Ek(total)=Ek+W= 120 J + 1800 J= 1920 J

So Ek(total)=1920 J = (1/2)*m*V² where V is the speed after the bicyclist increased his work output. So lets solve for V:

(1/2)*60*V²=1920

30*V²=1920, we divide by 30,

V²=64, and take the square root of both sides,

V=8 m/s.

So the speed of the bicyclist after the increased work output is V=8 m/s.

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Answer:

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3 years ago

A 750 kg car is stalled on an icy road during a snowstorm. A 1000 kg car traveling eastbound at 13 m/s collides with the rear of

scZoUnD [109]

(a) The magnitude of the velocity of the 1000 kg car after the collision is 10.5 m/s.

(b) The direction of the velocity of the 1000 kg car after the collision is 8.2 ⁰ north west.

(c) The ratio of the kinetic energy of the two cars just after the collision to that just before the collision is 0.72.

<h3>Velocity of the 1000 kg after the collision</h3>

Apply the principle of conservation of linear momentum as follows;

<h3>Final velocity in x direction</h3>

m₁u₁ + m₂u₂ = m₁v₁x + m₂v₂x

where;

m₁ is mass of 750 kg car
u₁ is initial velocity of 750 kg mass
m₂ is mass of 1000 kg car
u₂ is initial velocity of 1000 kg mass
v₁ is final velocity of 750 kg mass
v₂ is final velocity of 1000 kg mass

750(0) + 1000(13) = 750(4 cos 30) + 1000v₂x

13000 = 2,598.1 + 1000v₂x

10,401.9 = 1000v₂x

v₂x = 10.4 m/s

<h3>Final velocity in y direction</h3>

m₁u₁ + m₂u₂ = m₁v₁y + m₂v₂y

750(0) + 1000(0) = 750(4 sin 30) + 1000v₂y

0 = 1500 + 1000v₂y

v₂y = -1500/1000

v₂y = -1.5 m/s

<h3>Resultant final velocity</h3>

v = √(v₂ₓ² + v₂y²)

v = √[(10.4)² + (-1.5)²]

v = 10.5 m/s

<h3>Direction of the final velocity of 1000 kg car</h3>

tanθ = v₂y/v₂ₓ

tanθ = -1.5/10.4

tanθ = -0.144

θ = arc tan(-0.144)

θ = 8.2 ⁰ north west

<h3>Kinetic energy of the cars before the collision</h3>

K.Ei = 0.5m₁u₁² + 0.5m₂u₂²

K.Ei = 0.5(750)(0)² + 0.5(1000)(13)²

K.Ei = 84,500 J

<h3>Kinetic energy of the cars after the collision</h3>

K.Ef = 0.5(750)(4)² + 0.5(1000)(10.5)²

K.Ef = 61,125 J

<h3>Ratio of the kinetic energy</h3>

K.Ef/K.Ei = 61,125/84,500

K.Ef/K.Ei = 0.72

Thus, the magnitude of the velocity of the 1000 kg car after the collision is 10.5 m/s.

The direction of the velocity of the 1000 kg car after the collision is 8.2 ⁰ north west.

The ratio of the kinetic energy of the two cars just after the collision to that just before the collision is 0.72.

Learn more about kinetic energy here: brainly.com/question/25959744

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