Physics test questions

[t i l e x]

Hi guys. I've got my final exam on physics Friday and I guess I'm far from being ready so I'm stuyding a bit but there's a few exercises I can't manage to finish, I'll just post them here and if you have any idea it'll be more than welcome.

#5. A 80kg person pushes a 20kg box on a rugged surface. Consider us = 0.8 for the person and uc = 0.4 for the box. (a) What is the length of the maximum acceleration possible for the box ? (b) Use the answer to the previous question to find the length of the force applied by the box on the person.

The anwer to (a) is 27.4m/s^2 and (b) is 672N.

I'm pretty sure that the normals I've found are correct and they are 784N for the person and 196N for the box. Then, I don't know much what to do...

Any idea ?

[Alvaro]

The man pushing can only apply as much force on the box as friction he can get from the floor. That would be .8*80kg*9.8*m/s^2 = 627.2N. When that much force is applied on the box, the box will suffer an opposing friction of .4*20kg*9.8m/s^2 = 78.4N. So the net result is 548.8 N, which will accelerate the box at 27.44 m/s^2.

[t i l e x]

Thank you, but I'd like to know if you used any formula at all to find this, because in the test we have to write fornulas; I understand what you did but don't know how to write it so that it could get me all the points.

Here's another one that I missed...

#25. A 1.4 * 10^4 kg Polaris missile is propulsed by a 2 * 10^5 force. If its engine propulse it vertically for a minute starting from V = 0, how high will it go ? (we don't take account of friction nor the variation of g)

The answer is 11.8 km and I get 16.2 km :s

Here's what I did.

sum Fy = F - P = ma
F = ma + mg
2*10^5 = (1.4*10^4)a + (1.4*10^4)(9.8 )
4.49 = a

Vy = V0y + ay*t
Vy = 4.49 * 60
Vy = 269.4

Vy^2 = V0y^2 + 2ay*(y-y0)
Vy^2 = 2ay
269.4^2/4,49 = y
16,164m = y

It's probably a very simple error but I don't see where it is...

[Alvaro]

I agree that the acceleration is around 4.49m/s^2. Let's write acceleration, velocity and position as a function of time.
a(t) = 4.49
v(t) = 4.49 * t
y(t) = 4.49 * t^2/2

So the answer I get is y(60) = 8074 m, which doesn't agree with either of the numbers you gave.

I'm sorry I don't write down formulas. It goes against my religion.

[t i l e x]

The problem with your method using integrals is that v(t) = 4.49 * t + C where C is a constant which we don't know the value of. I guess I'll just ask my friend whenever she connects, she has an average of 99% in physics (she lost that 1% because she forgot to write down the units of an answer once...). Thank you for your help. I'll probably post more exercises, it's chapter 6 and there are 12 chapters...

[Alvaro]

The problem with your method using integrals is that v(t) = 4.49 * t + C where C is a constant which we don't know the value of.

But we do. The problem specifies "starting from V = 0", which means that C must be 0.

Feel free to post more exercises.

[t i l e x]

Here's another one, on work this time.

Consider a 1.8kg block moving at constant speed on a surface for which uc = 0.25. A force F is applied 45 degrees down of the horizontal axis and it moves over 2m. Find the work done on the block by (a) the force F; (b) the friction force; (c) the gravity force.

The answer are 11.8J, -11.8J (by Newton's law of intertia it could be guessed) and 0J.

This is what I have so far, I end up with 7.04J for the first answer.. :s

sum Fy = N - P - Fsin45 = 0
N = mg - Fsin45
N = 17.6 - Fsin45

sum Fx = Fcos45 - fc = 0
Fcos45 -0.25*(17.6 - Fsin45) = 0
Fcos45 + 0.25Fsin45 = 4.4
F = 4.98N

WF = Fscos theta
WF = (4.98 )(2)(cos 45)
WF = 7.04J

What's wrong ?

[Alvaro]

Fcos45 -0.25*(17.6 - Fsin45) = 0

Should be:
Fcos45 -0.25*(17.6 + Fsin45) = 0

The weight and the y component of the force both go down, so they should be added together.

[t i l e x]

Thanks a lot ! I forgot to switch the '-' to a '+' when I switched it to the other side of the equation. I guess I'll finish this chapter and it'll be enough for today (went from 4 to 7).

[t i l e x]

Here's another one:

The speed of a 2kg particle goes from (2i - 3j)m/s to (-5i + 2j)m/s. What is the variation of kinetic energy ?

The answer is 16.0J.

This is what I've done and I end up with 73.96J.

vf - vi = delta v
(-5i + 2j) - (2i - 3j)
(-5i + 2j) + (-2i + 3j)
(-7i + 5j) m/s

|v| = sqrt(7^2 + 5^2)
|v| = 8.60m/s

K = 0.5*m*v^2
K = (0.5)(2)(8.60^2)
K = 73.96J

[DannyBoy]

Ki=0.5*2*(sqrt(2^2+(-3)^2))^2=13J
Kf=0.5*2*(sqrt((-5)^2+(2)^2))^2=29J

Kf-Ki=29-13=16J

[t i l e x]

Thanks, you made me notice that it was on the formula sheet

[t i l e x]

Here's another one.

A propulsion force is opposed to a friction force of 250N. What is the work done by the propulsion force to move a 1000kg car in the following conditions: (a) at constant speed v = 20 m/s for 10s; (b) at constant acceleration from 0 to 20 m/s in 10s; (c) at constant acceleration from 20 to 40 m/s in 10s.

The answer is 5.0 * 10^4 J

I don't quite get what to do here.

[t i l e x]

Nevermind, I understood. It's Wtot = delta K - Wf and not K alone. I don't quite understand why it's delta K though, the speed is constant in (a)

In (b) I get 2.025 x 10^6J instead of 2.25 x 10^5 ?

[MXP]

a)
Since speed is constant the propulsion force equals the friction force. The speed and time tell you the distance that this force works over

d = vt
d = (20 m/s)(10 s)
d = 200m

W = Fd
W = (250N)(200m)
W = 50000 J = 5.0 x 10^4 J

b)
You know that there is an acceleration so there is a non-zero net force on the object. Drawing a free body diagram, you will get the following relation:

Fnet = M * a = Fp - Ffr (where Fp is the propulsion force and Ffr is the friction force)

(1000 kg)([20 m/s - 0 m/s] / 10 s) = Fp - 250 N
2000 N = Fp - 250 N
Fp = 2250 N

Now that you have the force, you just need to find the distance it works over.

d = (V0)(t) + (1/2)(a)(t^2)
d = (0 m/s)(10 s) + (1/2)(2 m/s^s)(10 s)^2
d = 100 m

W = (Fp)(d)
W = (2250 N)(100 m)
W = 225000 J = 2.25 x 10^5 J

You can use the same process as above to solve c).