[eal0088]

A 2.87 kg block is pushed 1.55 m up a vertical
Wall with constant speed by a constant force
Of magnitude F applied at an angle of 56.2◦
With the horizontal.
The acceleration of gravity is 9.8 m/s2 .
If the coefficient of kinetic friction between
The block and wall is 0.453, find the work done
By F.
Answer in units of J.

Find the work done by the force of gravity.
Answer in units of J.

Find the work done by the normal force be-
Tween the block and the wall.
Answer in units of J.

By how much does the gravitational potential
Energy increase during this motion?
Answer in units of J.

[Nhatkiem]

A 2.87 kg block is pushed 1.55 m up a vertical
wall with constant speed by a constant force
of magnitude F applied at an angle of 56.2◦
with the horizontal.
The acceleration of gravity is 9.8 m/s2 .
If the coefficient of kinetic friction between
the block and wall is 0.453, find the work done
by F.
Answer in units of J.

Find the work done by the force of gravity.
Answer in units of J.

Find the work done by the normal force be-
tween the block and the wall.
Answer in units of J.

By how much does the gravitational potential
energy increase during this motion?
Answer in units of J.

Wow that's a mouthful, lets get started.

A 2.87 kg block is pushed 1.55 m up a vertical
wall with constant speed by a constant force
of magnitude F applied at an angle of 56.2◦
with the horizontal...
If the coefficient of kinetic friction between
the block and wall is 0.453

m = 2.87, d=1.55m, theta = 56.2, mu=0.453

Okay, what we need to determine the force needed for constant velocity.

the block has a force going downwards = m*g, the forces pushing up are friction AND the force being applied at an angle :D!

F(g)=2.87*-9.8 = -28.13N
The forces counteracting this must be equal and opposite, so F(y)+F(friction) = -F(g)

F(y) = F*sin(52 deg)
F(friction) = mu*F*cos(52 deg) (the x component of the force exerted acts as the normal force)

F*sin(52)+0.453*F*cos(52)=28.13
F=28.13/[sin(52)+0.453*cos(52)]
F=26.4N

We now know all the forces. Work is just force * distance. I'll leave you to figure that out but If you have trouble post a reply and hopefully me or someone else will get to you :o

[eal0088]

Wow that's a mouthful, lets get started.

A 2.87 kg block is pushed 1.55 m up a vertical
wall with constant speed by a constant force
of magnitude F applied at an angle of 56.2◦
with the horizontal...
If the coefficient of kinetic friction between
the block and wall is 0.453

m = 2.87, d=1.55m, theta = 56.2, mu=0.453

Okay, what we need to determine the force needed for constant velocity.

the block has a force going downwards = m*g, the forces pushing up are friction AND the force being applied at an angle :D!

F(g)=2.87*-9.8 = -28.13N
The forces counteracting this must be equal and opposite, so F(y)+F(friction) = -F(g)

F(y) = F*sin(52 deg)
F(friction) = mu*F*cos(52 deg) (the x component of the force exerted acts as the normal force)

F*sin(52)+0.453*F*cos(52)=28.13
F=28.13/[sin(52)+0.453*cos(52)]
F=26.4N

We now know all the forces. Work is just force * distance. I'll leave you to figure that out but If you have trouble post a reply and hopefully me or someone else will get to you :o

Yea you totally lost me with what you've done. I'm having trouble discerning where you worked out each part of the question. Could you clear it up a bit for me?

[Nhatkiem]

Yea you totally lost me with what you've done. I'm having trouble discerning where you worked out each part of the question. Could you clear it up a bit for me?

Sure thing.

First I wanted to find out all the forces being exerted on this object. Because I knew the object was moving with a constant velocity, the sum of the forces must be zero, in other words, the forces in the y direction must be equal and opposite, and the forces in the x direction must be equal and opposite.

The y directional forces. These forces included gravity, the force of friction exerted on the crate, and the force positioned 52 deg from the floor. The reason I used a sine function was to find the y component of that force.

Gravitational Force = m*g
Force of friction = mu*F(normal)
F(y) = F*sin(52) , here F(y) stands for the y component of the force

In the x direction we have 2 forces, the force exerted by the force positioned 52 degrees from the floor (where I used a cosine function to determine the x component of the force) and the normal force the wall would have exerted on the box.

F(x) = F*cos(52) , here F(x) stands for the x component of the force
F(normal) = -F(x)

Our forces in the y direction had to be strong enough to counteract the force of gravity, but weak enough so that there was no acceleration. So they had to be equal and opposite.

Gravitational forces pulled down
Friction and the force applied countered the gravitational force.

-(Gravitational force) = friction + force applied , this value in the opposite direction as gravity

Now I had to determine the values for friction and the force applied in the Y direction.
From above we have

Force of friction = mu*F(normal)
mu is given, and the normal force is is -F(x) remember?
F(x) = , therefore

Force of friction = mu*F*cos(52)

F(y) = F*sin(52)

These two values added together must be equal to the gravitational force, but in the opposite direction.

-F(gravity) = F(friction) + F(y)
-m*g = mu*F*cos(52)+F*sin(52)

The only unknown here is F, the force being exerted at the 52 degree angle :D

Did that help?

[Nhatkiem]

Sorry, there must have been a mistake when I cut and pasted some text. it should say

"mu is given, and the normal force is is -F(x) remember?
F(x) = F*cos(52) , therefore.... "

Near the bottom of the page.

[eal0088]

It helps a bit, but it just looks to me that it's all answering one question and not the 4 individual parts.

Sorry if I'm putting you through the ringer with this.

[Nhatkiem]

Haha, no problem at all.

In truth I have not answered the 4 parts of the question, I left that for you to finish. What I have completed however was to find all the necessary information you'll need to complete those four parts.

If you look through the work, you'll see all the forces are accounted for. You just need to determine which of those forces are doing how much work

W = force * distance.

For example, the work done by the force of gravity would be

W = m*a*d
W = 2.87*9.8*1.55 = 43.6 J

again, you just need to determine what forces are doing what amount of work.

Before I forget, that last part, asking for potential energy can be found using

PE=m*g*h

PE is potential Energy, m is mass, g is gravity, and h is height.

[eal0088]

OK after reading through a few times it's starting to clear up. You said the only unknown left is F which is being exerted at 52 degrees. How would I go about finding that. I think this is what's stopping me right now.

[Nhatkiem]

The tricky part about this problem is that there are two forces pushing up, and 1 force pushing down.

As the force you apply (the one at 52 deg) increases, so does the frictional force.

I set the problem up in such a way so that the frictional force, and the y component of the force exerted were equal and opposite to the force of gravity.

Try looking again at the first and second reply.