Physics Question

[Duckers]

I saw this problem somewhere:

A 25 kg block is pushed accross a frictionless horizontal surface with a force of 20N, 20 degrees below the horizontal.
The Magnitude of the Normal Force of the block is:

a) 6.8N
b) 42N
c) 49N
d) 56N
e) 68N

I suck in Physics! I ended up with a huge answer.
I think that the answer might be (a), because 20sin20 = 6.84....
I am not sure if that's the right answer though. Me and some colleagues are debating whether the right answer was not included among the given choices.

 

[Napalm381]

Wellllllllll, there are three forces on the block: gravity (acts "down&quot, normal force (acts "up&quot and the force from the pulling, which can be decomposed into x and y components. The normal force must equal the sum of the y component of the pulling force and gravity, which would be 20*sin(20)+ 25*9.8= 6.84+245 =251.84. Hmmmmm, got me.

 

[Supradude]

specify your problem a little better, is it pushed or pulled at the 20 degree incline? all 3 forces probably have x/y components the way i'm seeing it now, if this is unanswered by the time i get hom today, i'll help ya again... but try to clarify your problem a bit...

 

[Napalm381]

I interpreted it being on a flat horizontal surface, not inclined, with the force acting on it at an angle.

 

[Skalman]

The right answer isn't included among the given choices.

 

[thEnEuRoMancER]

The answer is 238.4 N.

The block is not moving in the y direction and so the sum of forces in y direction must be 0. Gravity is acting down, y component of pushing force is acting up and so is the normal force. The force equation is:

Fg = Fy + Fn

Fn = Fg - Fy = m*g - F*sin20 = 25*9,81 - 20*0,342 = 238.4 N

Duckers, how did your 'bouncing ball' story end?
BTW, I still don't know how to determine elasticity of impact from k

 

[Pretender]

Nope...the answer is 20cos20

20sin20 would be the force at which it's accelerating down the slope, 20cos20 is the normal force.

[edit] I'm sure you use cos for the normal force when going down a slope, but the answer doesn't come out to one of the choices. Hmm, lemme contemplate this...I'll be back [/edit]


[edit #2] Another thing to mention...is gravity a factor in this equation? If so, you need to add the force of gravity also, that might do it [/edit #2]

 

[Pretender]

One more thing...is the block being pushed along the surface (20 degrees below the horizontal), or just completely horizontal?

[final edit until someone responds to my previous questions, since I'm confused ]I drew a simple diagram, x being the normal force we're looking for http://hcs.dhs.org/files/phys2.jpg

 

[thEnEuRoMancER]

Pretender - I understood this problem so that the angle between pushing force and the horizontal is 20deg and that the y component of the pushing force is pointing up and that the surface is horizontal.

BTW: check again your calculations. There is no way the y component of pushing force could be 20cos20 unless you place the angle in a very obscure way...