How Does Physics Explain These Real-World Scenarios?

[Affirmed]

Boredom...

I'm just plain stupid... (Don't know which equations to use...)

If a 256 kg spacecraft weights 436 Newtons when it is on the moon, what is the gravitational field strength (in N/kg) there?



To slide a 9.0 kg object at constant speed along the ground requires a force of 14.3 N. What is the coefficient of friction?



You are pulling a 50 kg mass over a rough surface of friction coefficient µ = 0.15 and you are pulling the mass using a spring of force constant = 200 N/m. How much would the spring stretch?



How much does a 55 kg girl compress the spring in a pogo stick when she stands on it? You are given that the spring constant is 78 N/cm.

 

[Curious3141]

You should post any ideas you have. I'm sure you've studied *some* equations, otherwise you wouldn't need to be doing these.

 

[Curious3141]

So what you're saying is that you're expected to solve all these questions without having ANY knowledge of the first thing about dynamics ? Because that's the impression you're trying to sell.

It's weak, dude. Unless you show that you've taken some effort, you'll get no help here.

 

[Affirmed]

If a 256 kg spacecraft weights 436 Newtons when it is on the moon, what is the gravitational field strength (in N/kg) there?

I would divide 436 N by 256 Kg = 1.703 N/Kg, unless you need to include the gravity for the moon.


To slide a 9.0 kg object at constant speed along the ground requires a force of 14.3 N. What is the coefficient of friction?

I would start off with µ = Fn/Ff
Fn = 9.0 * 9.8 = 88.2
Ff = 14.3 N

Then solve the equation as, µ = 88.2/14.3 = 6.17

As far for now...

$$\mu = \frac {FN} {Ff}$$

 

[Curious3141]

If a 256 kg spacecraft weights 436 Newtons when it is on the moon, what is the gravitational field strength (in N/kg) there?

I would divide 436 N by 256 Kg = 1.703 N/Kg, unless you need to include the gravity for the moon.

This answer and method is correct. You don't need to do anything more.

To slide a 9.0 kg object at constant speed along the ground requires a force of 14.3 N. What is the coefficient of friction?

I would start off with µ = Fn/Ff
Fn = 9.0 * 9.8 = 88.2
Ff = 14.3 N

Then solve the equation as, µ = 88.2/14.3 = 6.17

As far for now...

Your equation for $\mu$ is the wrong way around. Try solving it again, it should be less than one.

You are pulling a 50 kg mass over a rough surface of friction coefficient µ = 0.15 and you are pulling the mass using a spring of force constant = 200 N/m. How much would the spring stretch?

Use the same principles as in the previous question to find the kinetic frictional force resisting motion. Use Hooke's law $F = kx$ where k is given in the problem to determine $x$, the extension.

How much does a 55 kg girl compress the spring in a pogo stick when she stands on it? You are given that the spring constant is 78 N/cm.

What is the weight of the girl in Newton ? Use Hooke's law again.

 

[Affirmed]

To slide a 9.0 kg object at constant speed along the ground requires a force of 14.3 N. What is the coefficient of friction?

Yes I got it now, µ = Fn/Ff
µ = 14.3/88.2 = 0.16
(Thanks for telling me that the coefficient has to be less than 1)


You are pulling a 50 kg mass over a rough surface of friction coefficient µ = 0.15 and you are pulling the mass using a spring of force constant = 200 N/m. How much would the spring stretch?

Ff = µFn
Ff = (0.15)(490)
Ff = 73.5

F = kx
x = F/k
x = 73.5/200
x = 0.3675 m, 36.75 cm



How much does a 55 kg girl compress the spring in a pogo stick when she stands on it? You are given that the spring constant is 78 N/cm.

55 kg girl = 539 N
F = kx
x = F/k
x = 539/78
x = 6.91 cm

 

[Curious3141]

Yup, they all look right.

 

[HallsofIvy]

Yes I got it now, µ = Fn/Ff
µ = 14.3/88.2 = 0.16

You did the calculation correctly but the formula is μ= Ff/Fn.