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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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A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The
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Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

E_{i} = E_{f}

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}                              

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s      

 

3) When the object is at the equilibrium position, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s

I hope it helps you!                                                                                        

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