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###### A level maths mechanics question

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1 year ago

Why when a particle is deaccelerating on a horizontal surface that the driving force is 0. And if the driving force is 0, does that mean that if I find the resultant force it would be 0 - Friction on particle or Friction - 0?

(edited 1 year ago)

Original post by GTT21

Why when a particle is deaccelerating on a horizontal surface that the driving force is 0. And if the driving force is 0, does that mean that if I where to find the resultant force it would be 0 - Friction on particle or Friction - 0?

It generally helps to post the original question, but with the usual assumptions if there is a driving force acting on a particle on a horizontal table, then friction will act against the motion. The particle will decelerate when

resultant force = driving force - friction < 0

obviously this is satisfied when friction is non-zero and the driving force is zero.

Reply 4

1 year ago

When a particle is decelerating, the resultant force can’t be zero, according to Newton‘s second law - F=ma.

As there is no driving force, the friction must > 0 because there’s no other forces acting on the particle horizontally.

As there is no driving force, the friction must > 0 because there’s no other forces acting on the particle horizontally.

Original post by ChestnutCat

When a particle is decelerating, the resultant force can’t be zero, according to Newton‘s second law.

As there is no driving force, the friction must > 0 because there’s no other forces acting on the particle horizontally.

As there is no driving force, the friction must > 0 because there’s no other forces acting on the particle horizontally.

Original post by mqb2766

It generally helps to post the original question, but with the usual assumptions if there is a driving force acting on a particle on a horizontal table, then friction will act against the motion. The particle will decelerate when

resultant force = driving force - friction < 0

obviously this is satisfied when friction is non-zero and the driving force is zero.

resultant force = driving force - friction < 0

obviously this is satisfied when friction is non-zero and the driving force is zero.

Thanks for answering, I've attached the question I was referring to. They both deaccelerate and in both questions the driving force would be 0. But in part d of question 5, when finding the deceleration why do you do 0-T-resistance=ma and not T+Resistance-0=ma, does it depend on the direction of the acceleration?

(edited 1 year ago)

Original post by GTT21

Thanks for answering, I've attached the two questions I was referring to. They both deaccelerate and in both questions the driving force would be 0. But in part d of question 5, when finding the deceleration why do you do 0-T-resistance=ma and not T+Resistance-0=ma, does it depend on the direction of the acceleration?

The second question is "not found". For the first, assuming positive direction is right as per the initial driving force, then tension and resistance both act to the left and hence they're negative as would be the acceleration (deceleration). Note Ive not worked through the calcs to find the direction of the tension T, Im just assuming its the same direction as friction as you seem to imply.

If you assumed positive is left, then ma=T+resistance is indeed correct. However, the positive acceleration would be acting left and the initial velocity and resultant displacement would be negative as they are to the right.

(edited 1 year ago)

Original post by mqb2766

The second question is "not found". For the first, assuming positive direction is right as per the initial driving force, then tension and resistance both act to the left and hence they're negative as would be the acceleration (deceleration). Note Ive not worked through the calcs to find the direction of the tension T, Im just assuming its the same direction as friction as you seem to imply.

If you assumed positive is left, then ma=T+resistance is indeed correct. However, the positive acceleration would be acting left and the initial velocity and resultant displacement would be negative as they are to the right.

If you assumed positive is left, then ma=T+resistance is indeed correct. However, the positive acceleration would be acting left and the initial velocity and resultant displacement would be negative as they are to the right.

Thanks this is really clear! At first I put two questions up but it was easier to ask for help on one.

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