Answer :
Answer:
A) W=10J
B) [tex]\Delta K=10J[/tex]
C) [tex]K_{CM}=0J[/tex]
Explanation:
Since the problem is in one dimension, we can consider that one cart (1) is pushed in the positive direction while the other (2) is pushed in the negative one, so the work done on the system will be:
[tex]W=F_1.d_1+F_2.d_2=(5N)(1m)+(-5N)(-1m)=10J[/tex]
Using the work-energy theorem [tex]W_{Net}=\Delta K[/tex], and since all work done on the system is the one calculated before, we have [tex]\Delta K=10J[/tex]
The net force on the system is [tex]F_{Net}=0N[/tex], so there is no acceleration of the center of mass and it's kinetic energy is the initial one, [tex]K_{CM}=0J[/tex].
a) The work done on the two-cart system is 10J.
b) The change in kinetic energy of the system is 10J.
c) The kinetic energy of the center of mass of the system is 0.
Given the data in the question;
- Mass of cart; [tex]m = 0.5kg[/tex]
- Force pushing each cart; [tex]F = 5.0N[/tex]
- Distance; [tex]d = 1.0m[/tex]
a)
Work done on the two-cart system.
Work done measure of energy transfer that happens when an object is moved over a certain distance by an external force in the direction of the displacement. It is expressed as:
Work done = Force × Distance
[tex]W = F*d[/tex]
Now, work done on the two-cart system will be;
[tex]W = 2( F*d)[/tex]
We substitute in our value
[tex]W = 2( 5.0N * 1.0m)\\\\W = 2( 5.0N.m)\\\\W = 10N.m\\\\W = 10J[/tex]
Therefore, the work done on the two-cart system is 10J.
b)
The change in kinetic energy of the system.
From work energy theorem; the work done on a particle or object equals the change in its kinetic energy.
So,
Work done = change in kinetic energy
[tex]W = \delta K_E = 10J[/tex]
Therefore, the change in kinetic energy of the system is 10J.
c)
The kinetic energy of the center of mass of the system.
Since the two cart have the same mass, the forces exerted on the two masses are the same and they are moving with equal velocity; the velocity of the mass becomes zero(0).
Now, using the expression for Kinetic Energy:
[tex]K_E = \frac{1}{2}mv^2 \\[/tex]
Where m is mass and v is the velocity.
We substitute the value of velocity into the equation
[tex]K_E = \frac{1}{2}*m*(0)^2\\\\K_E = 0[/tex]
Therefore, the kinetic energy of the center of mass of the system is 0.
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