Physics force and work problem

In summary, the work done on the worker by the cake batter is -3822 J. This is calculated by considering the change in kinetic energy and gravitational potential energy of the worker as he falls through a total distance of 6.0 m. The energy lost to the batter is equal to the work done by the batter. The result can be improved by stating it to the correct number of significant figures.
  • #1
Jamest39
34
1

Homework Statement


A 65-kg worker at a bakery loses his balance and falls 4.0 m before hitting the surface of a large vat of cake batter. He continues to travel downwards an additional 2.0 m before the cake batter finally brings him to rest. Calculate the work done on the worker by the cake batter.

Homework Equations


F = ma
ΔKE = Wnet
W = Fscosθ
PEgravity = mgh

The Attempt at a Solution


ΔKE = KEfinal - KEinitial = 0 - KEinitial ⇒ KEintitial = -(1/2)mv^2
PEgravity = mgh = (65 kg)(9.80 m/s^2)(4.0 m) = 2548 J
 
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  • #2
Consider that the worker is still falling (moving vertically) through the cake batter until he comes to rest.
 
  • #3
gneill said:
Consider that the worker is still falling (moving vertically) through the cake batter until he comes to rest.

So,
F = ma = (65 kg)(9.80 m/s^2) = 637 N
W = Fscosθ = (637 N)(6.0 m)cos180° = -3822 J (6.0 as the displacement considering his whole movement?)
 
  • #4
That will work. The basic premise is that the work falls a total of 6 m through the gravitational field and so gains energy Mgh from the change in gravitational potential energy. That total amount of energy is "lost" to the cake batter over distance that he travels through the batter.
 
  • #5
gneill said:
That will work. The basic premise is that the work falls a total of 6 m through the gravitational field and so gains energy Mgh from the change in gravitational potential energy. That total amount of energy is "lost" to the cake batter over distance that he travels through the batter.

Was my answer right or do I still have to factor in the energy he lost while going through the batter?
 
  • #6
Jamest39 said:
Was my answer right or do I still have to factor in the energy he lost while going through the batter?
Your answer was fine. The energy lost to the batter is the energy gained from falling. That's why the worker comes to rest in the batter. That energy represents the work done by the batter. The only way you could improve your result would be to make sure that its stated to the correct number of significant figures.
 

Related to Physics force and work problem

1. What is the difference between force and work in physics?

Force is defined as a push or pull on an object that causes it to accelerate. On the other hand, work is the product of force and displacement, and it measures the energy transferred to an object by a force. In simpler terms, force causes an object to move, while work measures the amount of energy required to move an object.

2. How do you calculate the net force on an object?

To calculate the net force on an object, you need to add all the individual forces acting on the object. If the forces are acting in the same direction, you can simply add them together. However, if the forces are acting in different directions, you need to take into account their direction and magnitude using vector addition.

3. What is the relationship between force, mass, and acceleration?

According to Newton's Second Law of Motion, the force acting on an object is directly proportional to its mass and acceleration. In other words, the greater the mass of an object, the more force is needed to accelerate it. Similarly, the greater the acceleration, the more force is required to cause it.

4. How does work affect an object's kinetic energy?

Work is directly related to an object's kinetic energy. When work is done on an object, it gains kinetic energy, and when work is done by an object, it loses kinetic energy. This is because work is the transfer of energy, and kinetic energy is the energy an object possesses due to its motion.

5. Can a force be applied without doing any work?

Yes, a force can be applied without doing any work if the object does not move in the direction of the force. For example, if you push against a wall, you are applying a force, but the wall does not move, so no work is being done. Work is only done when there is a displacement in the direction of the force.

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