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kicyunya [14]
3 years ago
15

For each of the following cases, indicate whether the net work done is positive, negative, or zero. (a) gravity acting on a fall

ing ball (b) girl accelerating a wagon from rest (c) a skydiver falling at terminal velocity through the air (d) friction acting on a car so that it rolls to a stop.
Physics
1 answer:
irina [24]3 years ago
5 0

Answer:

Explanation:

We to indicate whether work is positive or negative or zero

Note, Work can be define as the produce of force and distance in the direction of the force, work is the dot product of force and displacement

W = F•d

(a) gravity acting on a falling ball

Work done acting on a falling ball is positive, since gravity does work on the ball. And g is positive downward.

Then, Work done by a free falling body is

W = mgh

(b) girl accelerating a wagon from rest

The workdone by a girl accelerating a wagon is positive

Since the wagon is a accelerating, then the velocity is increasing, then, the acceleration is positive

Work done by an accelerating car can be calculated be

W = F×d

Where the force is mass of the car times the positive acceleration

W = m•a•d

(c) a skydiver falling at terminal velocity through the air

The work done is zero since the terminal velocity is constant, so their is no change in velocity, the drag force is equal to the weight of the body and their is no external force acting on the body and the vertical acceleration goes to zero and since no acceleration, the work done is zero

W = m×a×h

W = 0J

But in this case, what is still don't by gravity. Since gravity is not zero.

(d) friction acting on a car so that it rolls to a stop.

The work done by friction is negative. This is due to the fact that the velocity of the car decreases with time and this implies that the car is decelerating, so the acceleration is negative, then,

W = m ×a × d

Since a =-ve

Then, W = —m•a•d

The work done by friction is negative

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Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

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You are the juror of a case involving a drunken driver whose 1041 kg sports car ran into a stationary 1928 kg station wagon stop
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1. Find the force of friction between the sports car and the station wagon stuck together and the road. The total mass m = 1928kg + 1041kg = 2969kg. The only force in the x-direction is friction: F = μ*N = μ * m * g 
2. Find the acceleration due to friction: 
F = m*a =  μ * m * g => a = μ * g = 0.6 * 9.81
3. Find the time it took the two cars stuck together to slide 12m:
x = 0.5*a*t² 
t = sqrt(2*x / a) = sqrt(2 * x / (μ * g) )
4. Find the initial velocity of the two cars:
v = a*t = μ * g * sqrt(2 * x / (μ * g) ) = sqrt( 2 * x * μ * g)
5. Use the initial velocity of the two cars combined to find the velocity of the sports car. Momentum must be conserved:

m₁ mass of sports car
v₁ velocity of sports car before the crash
m₂ mass of station wagon
v₂ velocity of station wagon before the crash = 0
v velocity after the crash

m₁*v₁ + m₂*v₂ = (m₁+m₂) * v = m₁*v₁ 
v₁ = (m₁+m₂) * v / m₁ = (m₁+m₂) * sqrt( 2 * x * μ * g) / m₁
v₁ = 33.9 m/s


7 0
3 years ago
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