Solving the Cart Friction Problem: Calculating Work Done by Force of Friction

[williamx11373]

a horizontal force of 200 Newtons is applied to a 55 kg cart across a 10 meter surface. If the cart accelerates at 2 meters/second squared, then what is the work done by the force of friction as it acts to retard the motion of the cart ??

Work= FX
work= maX

55 kg x 2 x 10 = 1100 - 200 N = 900 Joules ??


is this correct ?

 

[kuruman]

a horizontal force of 200 Newtons is applied to a 55 kg cart across a 10 meter surface. If the cart accelerates at 2 meters/second squared, then what is the work done by the force of friction as it acts to retard the motion of the cart ??

Work= FX
work= maX

55 kg x 2 x 10 = 1100 - 200 N = 900 Joules ??


is this correct ?

It is not correct. The "F" in F = ma is the net force, i.e. the sum of all the forces. In this case, it is the vector sum of the 200 N and the force of friction. What expression do you know that gives you the force of (kinetic) friction? Use that to calculate the work done by it.

 

[kerimek]

Yes, your calculation is correct. The work done by the force of friction in this scenario is 900 Joules. This means that 900 Joules of energy is being used to overcome the force of friction and keep the cart moving at a constant speed. It is important to consider the work done by the force of friction in situations like this, as it can affect the overall motion and energy of the system. By calculating this value, we can better understand the forces at play and make adjustments to optimize the motion of the cart.