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Tom [10]
3 years ago
5

Suppose you throw a baseball downward from a roof so that it initially has 120 J of gravitational potential energy, and 10 J of

kinetic energy. What will be true of the kinetic energy at the ground?
A.
It will be 0 J.
B.
It will be greater than 10 J.
C.
It will decrease from 10 J.
D.
It will equal 10 J.
Physics
2 answers:
castortr0y [4]3 years ago
8 0

Answer:

<h2>the correct answer is <u><em>B</em></u></h2>

Explanation:

Volgvan3 years ago
7 0

Answer:

B.

It will be greater than 10 J.

Explanation:

The total mechanical energy of an object is the sum of its potential energy (PE) and its kinetic energy (KE):

E = PE + KE

According to the law of conservation of energy, when there are no frictional forces on an object, its mechanical energy is conserved.

The potential energy PE is the energy due to the position of the object: the highest the object above the ground, the highest its PE.

The kinetic energy KE is the energy due to the motion of the object: the highest its speed, the largest its KE.

Here at the beginning, when it is at the top of the roof, the baseball has:

PE = 120 J

KE = 10 J

So the total energy is

E = 120 + 10 = 130 J

As the ball falls down, its potential energy decreases, since its height decreases; as a result, since the total energy must remain constant, its kinetic energy increases (as its speed increases).

Therefore, when the ball reaches the ground, its kinetic energy must be greater than 10 J.

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Read 2 more answers
Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li
Fofino [41]

Answer:

(a) t=3.87 s :time at which Kathy overtakes Stan

(b) d=37.36 m

(c) vf₁ = 15.097 m/s : Stan's final speed

    vf₂ = 19.31 m/s : Kathy's final speed

Explanation:

kinematic analysis

Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁: Stan displacement   

t₁ :  Stan time

v₀₁: Stan initial speed

vf₁: Stan final speed

a₁:  Stan acceleration

d₂: car displacement   

t₂ : Kathy time

v₀₂: Kathy initial speed

vf₂: Kathy final speed

a₂:  Kathy acceleration

Data

v₀₁ = 0

v₀₂ = 0

a₁ = 3.1 m/s²

a₂= 4.99 m/s²

t₁ = (t₂ +1) s

Problem development

By the time Kathy overtakes Stan, the two will have traveled the same distance:

d₁ = d₂

t₁ = (t₂ +1)

We aplpy the Formula (3)

d₁ = v₀₁t₁ + (1/2)*a₁*t₁²

d₁ = 0 + (1/2)*(3.1)*t₁²

d₁ =  1.55*t₁² ; Stan's cinematic equation 1

d₂ = v₀₂t₂ + (1/2)*a₂*t₂²

d₂ = 0 + (1/2)*(4.99)*t₂²

d₂ = 2.495* t₂² : Kathy's cinematic equation 2

d₁ = d₂

equation 1 = equation 2

1.55*t₁²  =  2.495* t₂²  , We replace t₁ = (t₂ +1)

1.55* (t₂ +1) ² = 2.495* t₂²

1.55* (t₂² +2t₂+1) = 2.495* t₂²

1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²

1.55t₂²+3.1t₂+1.55=2.495t₂²

(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0

0.905t₂² - 3.1t₂ - 1.55 = 0  Quadratic equation

Solving the quadratic equation we have:

(a) t₂ = 3.87 s : time at which Kathy overtakes Stan

(b) Distance in which Kathy catches Stan

we replace t₂ = 3.87 s in equation 2

d₂ = 2.495*( 3.87)²

d₂ = 37.36 m

(c) Speeds of both cars at the instant  Kathy overtakes Stan

We apply the Formula (1)

vf₁= v₀₁+a₁t₁    t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s

vf₁= 0+3.1* 4.87

vf₁ = 15.097 m/s : Stan's final speed

vf₂ = v₀₂+a₂ t₂  

vf₂ =0+4.99* 3.87

vf₂ = 19.31 m/s : Kathy's final speed

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