If the mass of the moon were tripled and it was moved to half as far f

[oneshot]

>>>If the mass of the moon were tripled and it was moved to half as far from the centre of the Earth as it now rests, what would the new gravitational force between the Earth and the moon be?


Can someone confirm with me that the answer is 12 times the original?


>> A 12 kg box is sitting on a ramp which is angled at 30deg
to the horizontal. If the coefficient of kinetic friction is 0.20, what is the acceleration of the box down the ramp? (Assume static friction is not high enough to prevent it from moving.)

I got an answer of 20.36N, can someone double check that for me too?

fk = uk Fn = (0.2)(101.8N) = 20.36N

Fn = (12)(9.8)cos 30 = 101.8N

 

[TSny]

Hi, oneshot. Welcome to PF.

Correct on the first question.

You still need to find the acceleration for the second question.

 

[oneshot]

oh right, thanks
Fnet is mg cos 30 - fk right
and then use Fnet = ma and solve for a

 

[TSny]

oh right, thanks
Fnet is mg cos 30 - fk right
and then use Fnet = ma and solve for a

Cos 30 is not correct for getting the component of the weight parallel to the ramp.

 

[Nickie]

To calculate the acceleration, we can use the equation Fnet = ma, where Fnet is the net force acting on the box and m is the mass of the box. In this case, the net force is the force of gravity pulling the box down the ramp, minus the force of friction pushing against the motion. So we can write:

Fnet = mgsin 30 - fk

= (12)(9.8)(sin 30) - 20.36

= 58.8 - 20.36

= 38.44 N

Now we can use Fnet = ma to find the acceleration:

38.44 = (12)a

a = 38.44/12

a = 3.2 m/s^2

So the acceleration of the box down the ramp is 3.2 m/s^2. This means that for every second the box is on the ramp, its velocity will increase by 3.2 m/s.