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GREYUIT [131]
3 years ago
9

It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp

ring an additional 0.10 m, does it take 130 J, more than 130 J or less than 130 J? Verify your answer with a calculation.
Physics
1 answer:
Ede4ka [16]3 years ago
6 0

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J

So, the new work is more than 130 J.

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A 150 g baseball is traveling horizontally at 50 m/s. If the ball takes 20 ms to stop once it is in contact with the catcher’s g
Sliva [168]
To solve for force, you need to get the product of mass and acceleration. 
F = ma

Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms

As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration. 
Acceleration is the change in velocity over a period of time.

a = change in velocity/time

To get the change in velocity, you get the difference between the initial velocity and final velocity:

a =  \frac{vf-vi}{t}

The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s. 

So we can put it into our formula now:

a = \frac{0m/s-50m/s}{20ms}

WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s. 

20ms x \frac{1s}{1000ms} = \frac{20s}{1000} = 0.02s

So your new time is 0.02s. Now we put this time into the formula:


a = \frac{0m/s-50m/s}{0.02s}
a =  \frac{-50m/s}{0.02s}  = -2,500 m/ s^{2}

As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop. 

So now we have our acceleration. Now using this, we can get our force. 

F= ma

Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is kg.m/ s^{2} but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first. 

150g =  \frac{1kg}{1,000g}  =  \frac{150kg}{1,000}  = 0.150kg

Our new mass is 0.150kg. 

To make things clearer, let us write down all our new values:

m = 0.150kg
a = -2,500 m/ s^{2}

Now that all our units match, we can put that into our formula:

F= ma
F= (0.150kg)(-2,500m/s^{2})
F = -375kg.m/ s^{2}  or -375N

The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign. 



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