Recent content by JustynSC

I see. In that case I will stick to my simple method :) Thanks!

Okay thanks I was thinking something along those lines, but why would the key ask to solve using that other equation? It seems more complicated than the way I did it. does my method only work in limited scenarios and the other has a broader range of uses?

Homework Statement A 10kg bucket of water is lifted vertically 3.0m at a constant speed. How much work did gravity do on the bucket during this process? Homework Equations Work=Force*Distance (what thought to use) Wext=Fextdcosθ (what the answer key says to use. The Attempt at a Solution My...

Hey thanks for that I have never heard of the phrase SUVAT equations. This was greatly helpful!

Oops. Yes the initial velocity is equal to 35.0m/s My frist attempt would have been a free body diagram depicting a ball traveling at the suggested speed coming in contact with the glove (or any object that acts against the ball's velocity. This in turn means there must be an opposing force...

Homework Statement A 0.140kg baseball traveling 35.0 m/s strikes the catcher's mitt, which in bringing the ball to rest, recoils backward 11.0cm. What is the average force applied by the ball on the glove Homework Equations F=ma; FAB=-FBA The Attempt at a Solution Not sure where to start for...

Awesome. Thank you for working that through without just telling me the answer. I think I actually understand the concept of how friction effects an object now.

Okay just to check my answer: fk=FN * μk Therefore: fk=98 N * 0.3μ Answer: fk= 29 N

(FN=ma=10kg * 9.8m/s2) So then, FN =98N ? P.S I am taking physics in college, but have had to teach myself because the professor does not do a great job at explain the reasoning behind the concepts. And his English is pretty limited, so there is a severe language barrier. So, sorry if I am...

At rest the box is acted on by Earth's gravity (9.8 m/s2) multiplied by the mass of the box (10kg)and lastly the static friction constant between the two surfaces (0.3) . So the Normal force is in the perpendicular to the surface the box is on. FN= 49 Newtons. But that is fs max, a threshold...

There is no normal force given. I have no knowledge on how many Newtons are being acted on the 10kg box. Is it required to have an FN given to solve that equation? Otherwise I only have one known variable, the 0.3 constant for static friction. I solved that the FN required to overcome static...

Homework Statement a 10 kg box is moving on a level floor. The coefficient of static friction between the box and floor is 0.3. What is the force of kinetic friction? Homework Equations fk = FN * μk The Attempt at a Solution I rearranged the formula so that everything was set equal to the...

Lucky for me I should be able to see them from my back porch right overhead! No need to drive anywhere, just sitback relax and gaze into the night sky

So I was looking into the problem a little further and I might have missed a minor detail that explains why they did not use +/- 0.05 as the error. But I still can't figure out how they justify the answer to be within +/- 0.2 Here is the Note at the start of all problems: assume a number like...

Thanks for the help! I get it now.