What edition is that book?

Wow! Thanks for the feedback. I guess I'll try to post specific questions...I'm drowning in assignments right now and am off to complete my chemistry which is due today. The physics instructor doesn't have due dates...that is good and bad.

here we go...

You are at the front of a floating canoe near a dock. You jump, expecting to land on the dock easily. Instead, you land in the water. Explain.

pennymiller said :

here we go...

You are at the front of a floating canoe near a dock. You jump, expecting to land on the dock easily. Instead, you land in the water. Explain.

For every action there is an opposite and equal reaction. Since you are jumping foward, you would expect the energy you generate that is transferred to the canoe to be transferred back to you which would propel you forward. But because the canoe is essentially fluid because it is not fixed or bound in one postion, the energy is actually tranferred into the water which actually provides more resistance than does the canoe.

My answer was in different words, but saying mostly the same thing?:

In this situation, you are at one with the canoe as a human/canoe system. The momentum of the human jumping from the canoe and the canoe moving in the water are equal in magnitude and opposite in direction. They cancel each other out. Pushing off the canoe = canoe moving away = fall into water below.

Here'a a harder one (expert level):

A bloc of ice of mass m is at rest atop an inclined plane of vertical height h. It then slides down the incline, reaching the floor below at speed v.

a) If friction can be ignored, show that the speed of the ice when it reaches the floor is the square root of 2gh.

b) If the block's mass is 27 kg and it slides down an incline of vertical height 1.5m, show that its speed at the bottom is 5.4 m/s.

a.)

  V_f² = V_i² +2gh

V_f² = 2gh

V_f = Sqr Root(2gh)

b.)

Using the same equations as above just plug the numbers in. (mass does not matter)

Vf = Sqrt(2*1.5*9.8) = 5.4

This is graduate level Physics?? I hope you are not a physics major...

As state above-- mass is irrelevant-- excluding friction-- any object with mass (on the macro scale) will accelerate at the same rate... don't get me started on quantum mechanics/statistical mechanics -- I never understood it, and stopped trying-- fortunately I could do the required math.... I was a chemistry Major-- Third year was Physical Chemistry (quantum and statistical mechanics)

I don't understand how some courses (such as physics) can be taught online?? Now courses like history/literature/etc... I understand...

I do wish you luck...

Manuel drops a water balloon of mass M from rest atop the roof of a building of unknown height. The balloon takes time t to hit the ground.

a) show that if air resistance is negligable, its kinetic energy just before it hits is 1/2mg^2t^2

b) If the water balloon has a mass of 1.2kg and the dropping time from rest is 2.0s, show that its kinetic energy when it hits the ground is 230 J

c) Why can't the force of impactbe found with the information given?

            a) KE = 1/2m x speed^2  or KE = (1/2m)(g^2t^2)

            b) KE = ½(1.2 kg)(10m/s^2)^2(2.0s^2)

                        KE = ½(1.2kg)(100)(4)

                        KE = 240 jouls

            The answer is supposed to be 230 joules…. ?  help!

            c) We do not know the distance. Net Force x distance = KE           

pennymiller said :

 

Manuel drops a water balloon of mass M from rest atop the roof of a building of unknown height. The balloon takes time t to hit the ground.

 

a) show that if air resistance is negligable, its kinetic energy just before it hits is 1/2mg^2t^2

 

b) If the water balloon has a mass of 1.2kg and the dropping time from rest is 2.0s, show that its kinetic energy when it hits the ground is 230 J

 

c) Why can't the force of impactbe found with the information given?

 

            a) KE = 1/2m x speed^2  or KE = (1/2m)(g^2t^2)

 

            b) KE = ½(1.2 kg)(10m/s^2)^2(2.0s^2)

                        KE = ½(1.2kg)(100)(4)

                        KE = 240 jouls

 

            The answer is supposed to be 230 joules…. ?  help!

 

            c) We do not know the distance. Net Force x distance = KE           

Could it be that g is actually 9.8 m/s^2?  If you use that value for g the answer becomes:

1/2*1.2*9.8*9.8*2*2 = 230.496 which is close to 230J.