hey guys need a little help in sec3 phy..hope u guys can help me..

1)a box of mass 20 kg is initially at rest when 2 boys A and B each apply a force of 45N and 35N respectively in the same direction on the box for a time of 10s

a)if the frictional force acting on the box is 20N for the first 10s, calculate the acceleration attained by the box in 10s.

b)if the frictional force acting on the box is increased to 80N after the first 10s, describle the subsequent motioon of the box in terms of its speed and acceleration.

Go see TYS lah.There are so many like this qns there

Originally posted by Martell21:

hey guys need a little help in sec3 phy..hope u guys can help me..

1)a box of mass 20 kg is initially at rest when 2 boys A and B each apply a force of 45N and 35N respectively in the same direction on the box for a time of 10s

a)if the frictional force acting on the box is 20N for the first 10s, calculate the acceleration attained by the box in 10s.

b)if the frictional force acting on the box is increased to 80N after the first 10s, describle the subsequent motioon of the box in terms of its speed and acceleration.

a) 3 m/s per s

b) the box will go from a velocity of 30 m/s to a stand still 0 m/s
The accelaration is -4 m/s per s

The box will travel 112.5 m before coming to a stand still and it will take 7.5 seconds for it to come to a stand still.

Originally posted by I_am_PeTe_Parker:

a) 3 m/s per s

b) the box will go from a velocity of 30 m/s to a stand
The accelaration is -4 m/s per s

The box will travel 112.5 m before coming to a stand still and it will take 7.5 seconds for it to come to a stand still.

hey thax alot but the working for b) is how ar??

b) the box will go from a velocity of 30 m/s to a stand still 0 m/s
The accelaration is -4 m/s per s

The box will travel 112.5 m before coming to a stand still and it will take 7.5 seconds for it to come to a stand still.

I doubt it will come to a stand still.. i believe it will instead continue moving at a constant velocity at 30 m/s with no acceleration because a net force 0N on the object doesn't mean that velocity = 0m/s

show me the workings pls

A Mothballz weighs 100N. What is the velocity of the Mothballz before it hits the ground when it is dropped from a height of 1m?

Originally posted by Martell21:

hey thax alot but the working for b) is how ar??

velocity at the start of part (b) = 0 + 3 X 10 = 30m/s (V = V0 + at)
Accelaration when friction is increase to 80N = 80N /20kg = -4 m/s per s

How long will it take for box to stop = (0 - 30) / -4 = 7.5 s
[t =(V - V0)/a]

How far will it travel = 30 X 7.5 + 0.5 X -4 X 7.5 X 7.5
= 225 - 112.5 = 112.5 m
[ x = V0t + 0.5 a X t X t ]

Originally posted by Mothballz:

A Mothballz weighs 100N. What is the velocity of the Mothballz before it hits the ground when it is dropped from a height of 1m?

I know this is a trick question....moth do not have balls...

Originally posted by Razor87:

I doubt it will come to a stand still.. i believe it will instead continue moving at a constant velocity at 30 m/s with no acceleration because a net force 0N on the object doesn't mean that velocity = 0m/s

But if you read the question again...you will notice the two boys only exert the force for 10s only. So in part b, which is already after 10s the only force acting is the friction of 80N.

But if you read the question again...you will notice the two boys only exert the force for 10s only. So in part b, which is already after 10s the only force acting is the friction of 80N.

oh right! I was caught!

Originally posted by Razor87:

oh right! I was caught!

Originally posted by Martell21:

hey guys need a little help in sec3 phy..hope u guys can help me..

1)a box of mass 20 kg is initially at rest when 2 boys A and B each apply a force of 45N and 35N respectively in the same direction on the box for a time of 10s

a)if the frictional force acting on the box is 20N for the first 10s, calculate the acceleration attained by the box in 10s.

b)if the frictional force acting on the box is increased to 80N after the first 10s, describle the subsequent motioon of the box in terms of its speed and acceleration.

I think the answer should be (the first answer was wrong).

(a)
F=ma
(45+35)-20=20a
a=20/60
a=0.333ms^(-2)

therefore, acceleration is 0.333ms^-2

The acceleration doesn't increase with time as the net force is constant across the first 10s.

(b)

v=u+at=0+0.333*10=3.33ms^-1

this is the velocity after the first 10s.

F=ma
(45-35)-80=20a
a=0

No net force therefore, the box will move at constant speed of 3.33ms^-1

Originally posted by findingnewidea:

I think the answer should be (the first answer was wrong).

(a)
F=ma
(45+35)-20=20a
a=20/60
a=0.333ms^(-2)

therefore, acceleration is 0.333ms^-2

The acceleration doesn't increase with time as the net force is constant across the first 10s.

(b)

v=u+at=0+0.333*10=3.33ms^-1

this is the velocity after the first 10s.

F=ma
(45-35)-80=20a
a=0

No net force therefore, the box will move at constant speed of 3.33ms^-1

this is correct anot ar???

Originally posted by I_am_PeTe_Parker:

But if you read the question again...you will notice the two boys only exert the force for 10s only. So in part b, which is already after 10s the only force acting is the friction of 80N.

this answer is correct, if it isn't a trick question as mentioned by peter parker - that the two boys still exert force on it after the 10s.

peter parker is correct

i assure u it`s rite
if and only if the 2 boys let go after 10 secs

thax alot guys

If I were you, I would make assumptions before the start of the question. First off, take gravitational force to be 9.81m/s. Next, assume frictional force to be negligible.

Originally posted by Martell21:

hey guys need a little help in sec3 phy..hope u guys can help me..

1)a box of mass 20 kg is initially at rest when 2 boys A and B each apply a force of 45N and 35N respectively in the same direction on the box for a time of 10s

a)if the frictional force acting on the box is 20N for the first 10s, calculate the acceleration attained by the box in 10s.

b)if the frictional force acting on the box is increased to 80N after the first 10s, describle the subsequent motioon of the box in terms of its speed and acceleration.

a) whats the meaning of acceleration in 10s? or does it mean accelation at time 10s ask the question setter to go and check his english unless he means to integrate the acceleration inmhich case it isn't acceleration already.

in addition, since the boys stop at 10 seconds and they ask for the acceleration at 10 seconds, i'm not sure but i seems that there is a discontinuity in acceleration since (lim t->10+)acceleration not equal (lim t->10-)acceleration. so the question may not have a single answer or it may the average of the 2 or somehting

i hope this question were nto set by you teacher or worse from the tenyear series man

wat sch are u from?

haha i think u betta go hire a tutor instead of asking ppl online like this...

if u want tutor... u can call mi =)

1hr 15$ =)

chey, thought what, sec 3 stuff.

dun need physics pros like me.

Originally posted by findingnewidea:

I think the answer should be (the first answer was wrong).

(a)
F=ma
(45+35)-20=20a
a=20/60
a=0.333ms^(-2)

therefore, acceleration is 0.333ms^-2

The acceleration doesn't increase with time as the net force is constant across the first 10s.

(b)

v=u+at=0+0.333*10=3.33ms^-1

this is the velocity after the first 10s.

F=ma
(45-35)-80=20a
a=0

No net force therefore, the box will move at constant speed of 3.33ms^-1

phys formula is correct.
but ur maths is wrong.
(45+35)-20=20a
80-20=20a
60=20a
20a=60
a=60/20
a=3

damn, careless enough. Luckily not in an exam.

thanks for the correction.