# Calc Push Force Req'd to Move 13g Steel Block on Steel Table

• bjbaldwi
In summary, to push a 13g block of steel across a steel table with a coefficient of friction of 0.6 at a steady speed of 1.2m/s for 8.8s, the work required is calculated by multiplying the friction force (0.6*13*9.8=76.44N) by the distance (1.2m/s*8.8s=10.56m), resulting in a total work of 807.2J. The force pushing the block (Fpush) is equal to the friction force (Ffriction) since the net force (Fnet) is equal to zero.
bjbaldwi
How much work must you do to push a 13g block of steel across a steel table(coefficent of friction=.6) at a steady speed of 1.2 for 8.8 ?

I know that Work=Force*Distance

i know what the D is but i don't know how to calculate the force.
i tried F=MA and got Fnet= Force pushed -Friction force
i really have no i dea tho. I need help please.

bjbaldwi said:
i tried F=MA and got Fnet= Force pushed -Friction force

How do you calculate the friction force?

Steady speed says the friction is constant and friction force is kinetic friction*normal force or Fk*mg mg= normal force i think

OK, good. But what does Fnet equal? (What's the acceleration of the block?)

sorry its 13 kg 1.2 m/s and 8.8 s

the acceleration is zero because it is a constant velocity

bjbaldwi said:
the acceleration is zero because it is a constant velocity
Good. So what's Fnet? And what does that tell you about Fpush?

Fnet is equal to zero since the acc. is zero. F push is equal to friction force?

bjbaldwi said:
Fnet is equal to zero since the acc. is zero. F push is equal to friction force?
Exactly! You should be able to calculate the work now.

Fpush is equal to 0 and f net is equal to friction force

so Fnet=Fpush+Ffriction
Fpush=0
Fnet=Ffriction
Ffriction=Fnet =(.6)mg---->.6*13*9.8=76.44
so
Work=Fnet*D
D = 1.2m/s*8.8s---->10.56 m
so
76.44*10.56=807.2 J

Thanks for all your help : )

bjbaldwi said:
Fpush is equal to 0 and f net is equal to friction force
No, you had it right before. Fnet = 0 so Fpush = Ffriction.

so Fnet=Fpush+Ffriction
Fpush=0
Fnet=Ffriction
No, but that doesn't affect your calculation, which is correct.

It's the force Fpush that is doing the work.

## 1. What is the formula for calculating the required push force to move a 13g steel block on a steel table?

The formula for calculating the required push force is F = μmg, where F is the force required, μ is the coefficient of friction between the steel block and table, m is the mass of the steel block, and g is the acceleration due to gravity (9.8 m/s^2).

## 2. How do I determine the coefficient of friction between the steel block and table?

The coefficient of friction can be determined by conducting a friction experiment, where the force required to move the steel block is measured at different weights and surface textures. The coefficient of friction is then calculated by dividing the force by the weight of the object.

## 3. Are there any other factors that may affect the required push force?

Yes, there are several other factors that may affect the required push force, such as the surface area of contact between the steel block and table, the angle at which the force is applied, and any external forces acting on the block.

## 4. Can the required push force be reduced?

Yes, the required push force can be reduced by increasing the coefficient of friction between the steel block and table, or by reducing the mass of the steel block. Additionally, using a lubricant between the two surfaces can also decrease the required force.

## 5. How is this calculation useful in real-world applications?

This calculation is useful in determining the amount of force required to move objects on different surfaces, which can be important in industries such as manufacturing, construction, and transportation. It can also aid in designing systems and equipment that require precise amounts of force for optimal functioning.

Replies
13
Views
2K
Replies
4
Views
880
Replies
61
Views
1K
Replies
7
Views
2K
Replies
8
Views
1K
Replies
3
Views
2K
Replies
3
Views
3K
Replies
2
Views
1K
Replies
3
Views
2K
Replies
1
Views
1K