Work and Energy

[Shay10825]

Hi everyone! I had these questions on a quiz I just took and I got them wrong. They will come back on the test so I need to understand how to do these.

1.) A 2 kg block slides down a frictionless incline from point A to point B. A force (magnitude P=3 n) acts on the block between A and B, as shown. Points A and B are 2 m apart. If the kinetic energy of the block at A is 10 J, what is the kinetic energy of the block at B?
http://img53.exs.cx/img53/5354/physics-quiz.jpg

2.) A .04 kg ball is thrown from the top of a 30 m tall building (point A) at an unknown angle above the horizontal. As shown in the figure, the ball attains a maximum height of 10 m above the top of the building before striking the ground at point B. If air resistance is negligible, what is the kinetic energy of the ball at B minus the kinetic energy of the ball at A?
http://img53.exs.cx/img53/2336/physics-quiz2.jpg

3.) A .40 kg particle moves under the influence of a single conservative force. At point A where the particle has a speed of 10 m/s, the potential energy associated with the conservative force is +40 J. As the particle moves from A to B, the force does +25 J of work on the particle. What is the value of the potential energy at point B?

4.) Starting from rest at t=0, a 2 kg block is pushed across a horizontal surface by a force as shown (P= 8N). The magnitude of the resulting acceleration of the block is 2 m/s^2. At what rate is the force of friction doing work on the block at t= 3 s?
http://img91.exs.cx/img91/711/Physics-quiz4.jpg

5.) A 10 kg block on a horizontal frictionless surface is attached to a light spring (force constant = .8 kN/m). The block is initially at rest at its equilibrium position when a force (magnitude P= 80 N) acting parallel to the surface is applied to the block, as shown. What is the speed of the block when it si 13 cm from its equilibrium position?
http://img91.exs.cx/img91/3631/physics-quiz3.jpg

Any help would be greatly appreciated.

[Pyrrhus]

Show us what you did for each problem.

so far, this is what i will say for each problem.

1)
Conservation of Mechanical Energy

\Delta K + \Delta \Omega = 0

2)
Conservation of Mechanical Energy

\Delta K + \Delta \Omega = 0

You could also help yourself with some kinematics.

3)
Work done by a conservative force

W = - \Delta \Omega

4)

Use definition of work

W = \vec{F} \cdot \vec{r}

Also help yourself with Newton's 2nd Law, and you could also use

\Delta E = W_{f}

5)
I was thinking a little more about this problem and you need

Work done by a conservative force [Spring Resistive Force]

W = - \Delta \Omega

And Work-Kinetic Enegry Theorem

\sum_{i=1}^{n} W_{i} = \Delta K

[Shay10825]

1.) F=3, m=2, d=2, Ka=10, Kb=?
Wother= U2-U1+K2-K1
Wother=K2-U1
(3)(2)=.5(2)(Vi^2)-.5(m)(vi^2)

How can I do this if I don't have the velocity?

2.) Wother = 0
-mgh=-(.04)(9.8)(30)=-11.76

3.) Vi=10, Ui=40, W=25, Uf=?

25= (Uf -40)+(.5mv^2)
25=(Uf-40)+[.5(.4)(100)]
Uf= 45

[Shay10825]

4.) P=W/t
how can I find the work?

5.) Vi=0, Vf=?, x=13, m=10

.5(m)(v^2)=.5(m)(kx^2)
.5(10)(Vf^2)=.5(.8)(13^2)
5(Vf^2)=.5(.8)(169)
Vf= 3.67

then I tried:

.5(m)(v^2)=.80
.5(10)(v^2)=.80
v=.4

I really did not know what to do for this one.

[Pyrrhus]

Heh, it's true what they say, if you do not understand the concepts, you won't know what to do.

I will check the other problems later, but on 5) You got two forces doing work on the block, and also read the work-kinetic energy theorem again.

[HallsofIvy]

4.) Starting from rest at t=0, a 2 kg block is pushed across a horizontal surface by a force as shown (P= 8N). The magnitude of the resulting acceleration of the block is 2 m/s^2. At what rate is the force of friction doing work on the block at t= 3 s?

Before you can use "P= W/t" you need to determine W!

If the acceleration of the 2kg block is 2 m/s2, then (F= ma), the total force must be 4 kg m/s2= 4 Newtons. But you are told that the force imposed is 8 N. What must the friction force be? Since that is a constant force, the work done by the friction force is just "force*distance". The "rate of doing work" is then that work divided by time: "force*distance/time" . But distance/time= speed. The rate of work done on the block by the friction is just the friction force (which you just found) times the speed. You are given the acceleration, what is the speed of the block after 3 seconds?