12 Maths based Physics problems

• Apr 21st 2007, 06:17 AM
nath_quam
12 Maths based Physics problems
Hey guys i have been having trouble with the following problems, any assistance would be much appreciated

Thanks Nath
• Apr 21st 2007, 07:10 AM
topsquark
Quote:

Question I
A physics textbook is suspended on a spring scale in an elevator. Of the following, the scale shows the highest reading when the elevator:
A moves downward with decreasing speed
B remains stationary
C moves upward at constant speed
D moves upward with decreasing speed
E moves downward with increasing speed

The scale is reading a value for the apparent weight of the object, aka the downward force acting on the spring. So the question is esentially asking about the force on the spring.

For the following I am defining upward to be the positive direction.

Let's take each of the answers and look at the value of the spring force.

A moves downward with decreasing speed
This means the acceleration is upward. So Newton's 2nd Law says:
SumF = F - w = + ma
(Where F is the spring force, w is the weight, etc.)

Solving for F gives us:
F = w + ma > w
So the spring force is greater than the weight.

B remains stationary
If the elevator is stationary, then there is no acceleration, thus:
SumF = F - w = m*0 = 0

F = w
So the spring force is equal to the weight.

C moves upward at constant speed
Again, there is no acceleration, so F = w as in case B.

D moves upward with decreasing speed
If the elevator is moving upward with decreasing speed the acceleration is downward, so:
SumF = F - w = -ma

F = w - ma < w
So the spring force is less than the weight.

E moves downward with increasing speed
Again the acceleration is downward, so F < w as in case D.

Putting it all together we see that answer A gives the largest spring force.

-Dan
• Apr 21st 2007, 09:31 AM
topsquark
Quote:

Question 2
A bullet, weighing 0.1 N, accelerates down a rifle of length 90 cm, reaching a speed
of 50 m/s as it leaves the gun. Assuming a constant acceleration as it travels through the rifle, the force acting on it is
A 0.57N
B 14.1N
C 28.4N
D 139N

It would appear that this is a Newton's 2nd Law problem, except that we don't know the acceleration to find the force. So how do you find a constant acceleration?

Let the back of the barrel of the gun be the origin and let the +x direction be along the barrel. We know that:
x0 = 0 m
x = 0.90 m (The bullet accelerates until it leaves the barrel.)
v0 = 0 m/s
v = 50 m/s
a = ?
and we don't know t, the time it takes the bullet to leave the barrel.

It looks like we can use
v^2 = v0^2 + 2a(x - x0)

a = (v^2 - v0^2)/[2(x - x0)]

a = v^2/(2x) <-- After inputting the 0 values

a = (50 m/s)^2/(2*0.90 m) = 1388.89 m/s^2

So to find the net force on the bullet we simply do
F = m*a.

BUT!! Have a care! Many students tend to make a mistake here and try to use 0.1 N as the mass. 0.1 N is a weight, meaning the mass of the bullet can be found by:
w = mg

m = w/g = (0.1 N)/(9.8 m/s^2) = 0.010204 kg

So the net force on the bullet would be:
F = (0.010204 kg)*(1388.89 m/s^2) = 14.1723 N

-Dan
• Apr 21st 2007, 10:11 AM
topsquark
Quote:

Question 3
A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. When the spring is 4.0 cm shorter than its equilibrium length, the speed of the block is 0.50 m/s. The greatest speed of the block is:
A 0.23 m/s
B 0.32 m/s
C 0.55 m/s
D 0.71 m/s
E 0.93 m/s

This is an energy question.

The total mechanical energy of the spring when the spring is 0.04 m shorter than the equilibrium length is:
S = (1/2)kx^2 = (1/2)*(80 N/m)*(-0.04 m)^2 = 0.064 J.

At this spring compression the block contains an additional kinetic energy of
K = (1/2)mv^2 = (1/2)*(0.50 kg)*(0.50 m/s)^2
= 0.0625 J

So the total mechanical energy in the system is
E = S + K = 0.0064 J + 0.0625 J = 0.1265 J

The maximum speed of the block is when all of the energy of the system is stored entirely in the kinetic energy. So:
E = K' = (1/2)mv'^2
(where K' is the new kinetic energy of the block and v' is the maximum speed of the block.)

v' = sqrt{2K'/m} = sqrt{2*(0.1265 J)/(0.50 kg)} = 0.711337 m/s

So the maximum speed of the block is 0.71 m/s, or answer D.

-Dan
• Apr 21st 2007, 11:46 PM
nath_quam
4 B
5 A
6 A
7 C OR E
8 C
9 E
10 D

are they correct and for 11 and 12 i wasn't sure
• Apr 22nd 2007, 05:54 AM
topsquark
Quote:

Originally Posted by nath_quam
4 B
5 A
6 A
7 C OR E
8 C
9 E
10 D

are they correct and for 11 and 12 i wasn't sure

4 A
The amount of work done by the force F in stopping both blocks is the same since both blocks initially had the same kinetic energy. (W = (Delta)K) Since W = Fd and the force applied is the same in both cases, the stopping distance must be the same.

7 E
By definition the total energy of a system E is the sum of all potential and kinetic energies of the system. If the system is closed (ie. nothing enters or leaves the system) then E is conserved. Since this answer isn't listed, the correct answer is E.

11 B
The short version is that a = g*sin(theta) where theta is the angle of inclination. I don't have time right now to write out how to get this solution, so please make sure you look up how to do it!

12 C
The (negative) change in gravitational potential energy is equal to the work done by the weight. This is equal to W = -(Delta)PE = mg(Delta)h. Since (Delta)h = 0 m, the work done by the weight is equal to 0 J.

Work done by ANY resistive force implies the force is in the opposite direction to the (infinitesimal) displacement at that point. Thus dW = F (dot) ds is always negative, so W = Int[dW] is always negative.

-Dan
• Apr 22nd 2007, 06:00 AM
nath_quam
Thanks
Just quickly does that mean the other answers where correct?

Nath