Work problem of person pushing box

[jacy]

hi,
Please help me with this problem, thanks.

Harry pushes a box (that weighs 98 N) with a force of 20 N on ice. After he stops pushing the box is still in motion. The kinetic energy of the box is 20 J and the coefficient of friction is .05. Find
a) Mass of the box?
b) Force of friction on the box?
c) Magnitude of work done by friction to stop the box?
d) How far the box will slide before stopping?

I figured out the 1st part.
F= mg
98 N = m * 9.8 m/s^2
m= 10 kg

The mass of the box is 10 kg.

I don't know how to proceed further. Please help, thanks.

 

[andrewchang]

Part B can be solved using the coefficient of friction and the normal force.
Part C would be the same as the kinetic energy, because all of the energy must be dissipated.
Part D would be solved using the Law of Conservation of Energy.

 

[jacy]

Part B can be solved using the coefficient of friction and the normal force.
Part C would be the same as the kinetic energy, because all of the energy must be dissipated.
Part D would be solved using the Law of Conservation of Energy.

This is what am coming up for the B part

Fn - mg -Fp sin(theta) = 0
Fn = mg + Fp sin(theta)

mgcos(theta) - Ff = ma

I don't know the angle theta, what should i do know, thanks.

 

[Jeann25]

I don't see any angle in the original problem you posted. According to the problem, it sounds like it is just being pushed straight, so no angle should be involved. And you don't need the mgcos(theta) - Ff = ma for part B. To solve for friction you only need the normal force and the coefficient of friction.

 

[Izekid]

I have solved the problem for u, For nr 2 You take Fg*m*0,05 = And you get the Frictionwork. For nr 3 its the same because all energy does not go away it transforms so it must be 20J. And nr 4 it's just to take S = W/F because the definition for work is W= F*S

I hope this is right, this is what is what popped up for me anyway.

 

[jacy]

I have solved the problem for u, For nr 2 You take Fg*m*0,05 = And you get the Frictionwork. For nr 3 its the same because all energy does not go away it transforms so it must be 20J. And nr 4 it's just to take S = W/F because the definition for work is W= F*S

I hope this is right, this is what is what popped up for me anyway.

Thanks, but won't u think the friction force and displacement are in the opposite direction so the angle between them will be 180 degrees.
W= F*d cos 180. What do u think, thanks for looking at my problem.

 

[jacy]

I don't see any angle in the original problem you posted. According to the problem, it sounds like it is just being pushed straight, so no angle should be involved. And you don't need the mgcos(theta) - Ff = ma for part B. To solve for friction you only need the normal force and the coefficient of friction.

Thanks for your help.

Ff = coeff of friction * Fn
= .05 * 98 N
= 4.9 N

For the work done by friction force
W(f) = Ff * d cos 180
since the friction and displacement are in the opposite direction,so the angle between them is 180. Now what should i do for W(f). I have two unknowns W(f) and d, thanks.

 

[NateTG]

I have two unknowns W(f) and d, thanks.

Actually, you can figure out what W(f) is from the problem.

 

[jacy]

Actually, you can figure out what W(f) is from the problem.

Will W(f) will be -Ff, thanks

 

[NateTG]

The box starts out in motion (with 20 Joules of Kinetic Energy) and then comes to a stop. What forces are doing work on the box?

 

[andrewchang]

drawing a free body diagram might help

 

[jacy]

The box starts out in motion (with 20 Joules of Kinetic Energy) and then comes to a stop. What forces are doing work on the box?

The forces doing work on the box are Ff (frictional force), Fn (normal force), gravity (mg), and Fp (force of pus). When the box is in motion it has kinetic energy. Am not getting this one sorry.

 

[andrewchang]

is there really the force of the push on the box? check the question again.

 

[lightgrav]

Force describes a *condition*, visible in a snapshot diagram (FBD).
Work is something that occurs (is DONE) during a process -
the process changes the starting condition into the ending condition.

The Force component *parallel to* the displacement does Work
as the displacement is accomplished (W = F.dx)
This means that Energy is transferred from one object to another
(thru the contact area) as that contact point moves.

Where does the block's KE end up?

 

[jacy]

is there really the force of the push on the box? check the question again.

The force of push is there on the box intially. After the box begins to move there is no push.

 

[jacy]

Force describes a *condition*, visible in a snapshot diagram (FBD).
Work is something that occurs (is DONE) during a process -
the process changes the starting condition into the ending condition.

The Force component *parallel to* the displacement does Work
as the displacement is accomplished (W = F.dx)
This means that Energy is transferred from one object to another
(thru the contact area) as that contact point moves.

Where does the block's KE end up?

The box's KE ends up when it comes to stop.

 

[andrewchang]

The force of push is there on the box intially. After the box begins to move there is no push.

yeah, so that means that there's 20 Joules of kinetic energy initially, and the work done by friction decreases it all to zero. So, the -W_f = KE. You can solve for the displacement of the box by using W_f = F d cos(Theta)

the direction doesn't matter, since the question asks for a magnitude

 

[jacy]

yeah, so that means that there's 20 Joules of kinetic energy initially, and the work done by friction decreases it all to zero. So, the -W_f = KE. You can solve for the displacement of the box by using W_f = F d cos(Theta)

the direction doesn't matter, since the question asks for a magnitude

Thanks for ur help.