Recent content by hkor

Hey, i have the same question to answer. To find the initial velocity of the ball do we only consider the horizontal velocity or must we also find the vertical velocity? If so, how do we find it ?

ok so if F(s)= mg F(s) =(2)(9.8) = 19.6N Fs= 0.6xn n= 19.6/0.6 n= 32.67 In order for the box to stick to the cart the force going in the positive x direction must be equal to or greater than the normal working in the -x direction? F=n=ma 32.67=(2)(a) a=16.335m/s/s?

Okay i think I've got it. well the answer is right anyway. I think if we assume that the spring is compressed all the way to the bottom of the ramp, transferring all of its kinetic and potential energy to the elastic potential of the spring. PE(final) =0, KE(final)=0 mgh(initial) +...

Sorry i don't really understand. I've looked at a couple of workings of the problem and the only thing I can't figure out is how to find the height. The distance of the box is initially at the 3m from the spring. But we don't know the length of the spring to the bottom of the ramp? Is the...

So the friction force is acting upwards and the weight force is acting down? therefore (2kg)(9.8N) = Fs Is the normal going in the horizontal direction?

Okay so at the initial position the box has PE= mgh and KE = 0.5mv^2. At its final position the box is a rest therefore KE = 0 and all energy is transferred into the PE of the spring? using the elastic potential energy of a spring mgh + 1/2mv^2 = 1/2kx^2 + 0 (2.5kg)(9.8N)(h)...

Sorry for the confusion the mass of the box is 2 kg and the box is in front of the cart :) I guess I still don't really understand the concept. If the static friction of the box is acting downwards then why is it dependent on the force to the right whether the box stays stuck to the cart...

conservation of energy Do you mean mgh(initial) +0.5mv^2(initial)= mgh(final) +0.5mv^2 where the initial velocity of the box is 0.75, the initial height .3m and the final velocity of the box zero, h final is the only unknown?

Homework Statement A frictionless, incline plane at 20 degrees has a spring with force constant k=500N/m attached to the bottom, parallel to the inclined plane. A block of mass m=2.5kg is placed on the plane at a distance d =3m from the spring. From this position the block is projected...

attempt at answering the question Actually I do understand, now that i think about it. So when the acceleration is greater than or equal to the static force we know that the box will stick to the cart.? So this is what I've come up with.. F(y)=n-mg=0 therefore n-mg mg=(2.5x9.8) =24.5...

Homework Statement You place a 2kg box against the side of a cart. Nothing is used to stick the box to the cart which accelerates to the right. The coefficient of static friction is 0.6. Find the minimum acceleration of the cart needed so that the box does not fall Homework Equations...

Homework Statement A physicist pours 100g of water at 100'C into a 20g aluminium cup which already contains 50g of water at 20'C. The final equilibrium temperature of the water is 71.847. C(water)=4186J/kgC C(aluminium)=900J/kgC 1. the amount of heat lost by the hot water 2. the amount...

So when finding the dot product i use the unit vector (2/sqrt(14),3/sqrt(14),-1/sqrt(14)) . (1,0,0) = 2/sqrt(14). The magnitude of the normal vectors will be one, thus the ans will be... cos ^-1 (2/sqrt(14))= 1.0069?

Oh great thanks for the hint. I wasn't aware that for the angle between intersecting planes, the normal vector was used, unlike in the angle between intersecting lines. I think this is correct - the normal vector of 2x+3y-z=4 is (2,3,-1) and the normal vector of x=0 is (1). Thus the dot product...

how do i find the angle between the plane x=0 and the plane 2x+3y-z=4?