Work transfer and energy conservation

In summary, the gardener has to give an extra force to keep the lawnmower going, and the work done by the frictional force is 'lost'.
  • #1
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1)A gardener pushes a 12 kg lawnmower whose handle is tilted up 37 degrees above horizontal. The lawnmower's coefficient of rolling friction is 0.15. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.2 m/s?

not really sure where to begin with this one. so any suggestions as to how to start would help.

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The spring in the figure has a spring constant of 1200 N/m. It is compressed 15.0 cm, then launches a 210 g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.210. What distance d does the block sail through the air?

http://s93755476.onlinehome.us/stuff/knight.Figure.11.72.jpg

this one seems like it has several parts. i know i could figure out the velocity that the block travels on the flat surface from the spring using like K + U = K + U for initial and final. and using 1/2kx^2 instead. and something simliar for going up the hill. but the part where its in the air is where i get lost.
 
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  • #2
1) Calculate the normal force on the mower to get the frictional force.
The 'extra' force the gardener has to give to keep the mower going is equal and opposite to this frictional force.

2) Calculate the the work done by the frictional force when the block slides up the hill.
This energy is 'lost'. Now you the energy of the block at the end of the hill it becomes a trajectory problem.
 
  • #3
i am lost on what to do for the first question about the lawnmower..can someone help me?
 
  • #4
Big hints in lawnmower question:

Constant velocity, what do you know about the net force on objects with constant velocity?
The force of friction plays an important role here. You can find the force of fricion by the information in the problem, use that force and the hint above this to find the force applied to the lawnmower.
Knowing the force on the lawnmower, find the force applied by the gardener, considering the angle he is pushing.

From there you have a force and a velocity. Power = Force x Velocity
 

FAQ: Work transfer and energy conservation

What is work transfer?

Work transfer refers to the movement of energy from one system to another. This can occur through various means such as mechanical work, electrical work, or thermal work.

How is work transfer related to energy conservation?

Work transfer is an important component of energy conservation. According to the law of conservation of energy, energy cannot be created or destroyed, only transferred or converted from one form to another. Therefore, any work transfer must be accounted for in the conservation of energy within a system.

Can work be transferred without any energy loss?

In theory, yes. Work can be transferred without any energy loss if the transfer is carried out in a perfectly efficient manner. However, in reality, there will always be some energy loss due to factors such as friction, heat transfer, and other forms of energy dissipation.

How is work transfer calculated?

The amount of work transferred is calculated by multiplying the force applied by the distance over which the force is applied. This can be represented by the formula W = F x d, where W is work, F is force, and d is distance.

What are some examples of work transfer in daily life?

Some common examples of work transfer in daily life include pushing a car, lifting a heavy object, and turning on a light switch. These actions all involve the transfer of energy from one system to another to perform work.

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