Change in kinetic energy of a box

In summary, when a 41 kg box is pushed 4 m along a rough, horizontal floor with a constant force of 121 N, the work done by the applied force is 4.84e2 J. The increase in internal energy of the box and floor system due to friction is 4.66e2 J. The work done by the normal force and gravitational force on the box are both 0 J. The sum of external work is 4.84e2 J and the change in kinetic energy of the box is 4.66e2 J.
  • #1
bearhug
79
0
A 41 kg box initially at rest is pushed 4 m along a rough, horizontal floor with a constant applied force of 121 N. If the coefficient of kinetic friction between the box and the floor is 0.29, find
(a) the work done by the applied force W= FΔx
Ans: 4.84e2 J
b) the increase in internal energy of the box + floor system due to friction ΔEint= fkd fk= ukn= 116.5N
ΔEint= 4.66e2 J
c)the work done by the normal force on the block due to the floor
=0.00 J
d) the work done by the gravitation force on the block due to the Earth =0.00 J
e)the change in kinetic energy of the box
ΔK = -fkd + ∑W This is where I'm stuck. Is ∑W= to 4.84e2J because I thought that the applied force was included in the change in kinetic energy but this according to my homework is wrong. Can someone explain what I'm missing here?
 
Physics news on Phys.org
  • #2
Thanks! The sum of external work (ΣW) is the sum of all forces applied by external agents, i.e. the applied force (F=121N), the normal force (fn) and the gravitational force (mg). In this case, Wapp = FΔx = 4.84e2 J, Wnormal = 0 J and Wgravity = 0 J, so ΣW = 4.84e2 J. The change in kinetic energy of the box is then ΔK = -fkd + ΣW = -(0.29)(41)(4) + 4.84e2 J = 4.66e2 J.
 
  • #3


The change in kinetic energy of the box can be calculated by subtracting the work done by friction and the net work done by all other forces (such as the applied force and the normal force) from the total work done by the applied force (as you have correctly calculated in part a). In this case, the change in kinetic energy would be equal to the negative of the work done by friction (since friction is acting in the opposite direction of motion) plus the net work done by all other forces. So, the equation would be: ΔK = -ΔEint + ∑W. Plugging in the values from parts b and c, we get: ΔK = -4.66e2 J + 4.84e2 J = 1.8e1 J. This means that the change in kinetic energy of the box is an increase of 18 J.
 

1. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is a type of energy that is dependent on an object's mass and velocity.

2. How is kinetic energy calculated?

The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

3. What causes a change in kinetic energy?

A change in kinetic energy can be caused by a change in an object's mass, velocity, or both. An increase in either of these factors will result in an increase in kinetic energy, while a decrease will result in a decrease in kinetic energy.

4. How does a box's mass affect its kinetic energy?

The kinetic energy of a box is directly proportional to its mass. This means that the greater the mass of the box, the greater its kinetic energy will be, assuming the velocity remains constant.

5. How does friction affect the change in kinetic energy of a box?

Friction is a force that acts against the motion of an object. When a box is moving, friction will cause a decrease in its kinetic energy by converting it into other forms of energy, such as heat. This is why it is important to minimize friction in certain situations, such as in sports or transportation, in order to maintain a higher kinetic energy.

Similar threads

Replies
1
Views
460
  • Introductory Physics Homework Help
Replies
5
Views
714
  • Introductory Physics Homework Help
Replies
33
Views
815
  • Introductory Physics Homework Help
Replies
7
Views
934
  • Introductory Physics Homework Help
Replies
2
Views
149
  • Introductory Physics Homework Help
Replies
14
Views
952
  • Introductory Physics Homework Help
Replies
15
Views
276
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
6K
Back
Top