Recent content by NewJersey

Fs1=1/2 kx^2 so if k = 25 v=4 = 200 Fs2= k=50 v=4 =400 so by factor of 2

An elevator is running with a 25hp motor. to double the lift speed a ____ hp motor is required. To double the lift speed should we just times 25 by 2 and get 50hp motor. Or is there something more to this.

If compression of a spring is doubled, the stored energy increased by a factor of? I think that the stored energy is increased by a factor of 2. Right?

Ok , what about this equation 24J= 12cos(degree) can i set it up like this to find the answer , if so is the answer 66degrees.

ok, i found the max height already, and it was 7.3. So I think you are saying that using that i can find the angle b , using what equation?

ok, this is what i figured out so far mgh=1/2mv^2 which m can cancel out gives gh-1/2v^2 which can give me h which is h= v^2/2g which I can solve for is 12^2/ 2*9.87m/s = 7.3m is this right?

A .050-kg ball is shot at a velocity of 12 m/s by an upward-slanting cannon. At its maximun height, the kinetic energy of thew ball is 24J. a) Tha max. height is? b) the angel of the cannon to the horizontal is? I really don't know where to start on this question. If someone can...

I was following a homework example, I must have wrote down N as a mistake.

ok, Yeah I just saw the equation in my notes.

so for part b can I used the equation h=Kx^2/2mg 2500N*M * (.080)^2 / 2*.25kg*9.87m/s = 19.74m

Okay I see what you are saying I found the force instead of work? 1/2*2500N/M*(.060)^2= 4.50J 1/2*2500N/m*(.080)^2 -4.50J= 3.5J Now how do I find the max height, I am confused about that part

A spring having constant of 2500N/m is compressed 6.0cm and then additional 2.cm. Then the compressed spring is used to propel a 0.25-kg ball vertically upward. a) The additional work done on the spring in compressing te final 2.0cm was ? b) the maximun heignt of the ball was part...

The second part has friction and the question is In the question above, if the coefficient of friction is 0.250 a) The work done on the box was? b)the final velocity of the box was ? And I set it up like part a = 3kg*9.87*sin(20) = 10.05 3kg*9.87*cos(20)= 27.62 Un=...

Okay, I am sorry for the confusion. This is a two part question the first part is frictionless, and it states.. A 3.00kg box, initially at rest, slides 1.50m down a frictionless plane inclined at 20degrees to the horizontal? a) The work done on the box was ? B) the final velocity of...

Ok in the first question , you said used the equation mg*sin(20)-(mu)mg*cos(20). And mu=frictional force. However in the first equation we are to asumed there is no frictional force because it is no stated. So this equation is used for the second question where frictional force is given?