How Much Work Does a Worker Do Pushing a Block on a Frictional Surface?

[KMjuniormint5]

Homework Statement

A worker pushed a 27 kg block 7.8 m along a level floor at constant speed with a force directed 32° below the horizontal.
(a) If the coefficient of kinetic friction is 0.20, how much work was done by the worker's force?
(b) What was the increase in thermal energy of the block-floor system?

Homework Equations

Probably could use Fnet = ma

Work = F*distance

Friction is \muN

The Attempt at a Solution

(a) If the coefficient of kinetic friction is 0.20, how much work was done by the worker's force?

I drew a FBD and found N = (Weight)+(Force of worker in Y-axis(Fwy)) and in the x direction (Fwx - Friction force) = ma

Fwx - \mu[(Weight)+(Force of worker in Y-axis(Fwy))] = ma

would a = 0m/s^2 in the x-direction since the object is moving at a constant speed. . .if yes then. . .

Fwx = \mu[(Weight)+(Force of worker in Y-axis(Fwy))]

but then I get lost again. . . .

 

[kamerling]

Homework Statement

A worker pushed a 27 kg block 7.8 m along a level floor at constant speed with a force directed 32° below the horizontal.
(a) If the coefficient of kinetic friction is 0.20, how much work was done by the worker's force?
(b) What was the increase in thermal energy of the block-floor system?

Fwx = \mu[(Weight)+(Force of worker in Y-axis(Fwy))]

This is OK. You have everything you need to calculate Fwx. (I would call Fwy "Force of worker in Y direction")
The work done is just (force * distance) for a constant force pushing in the direction of movement.

 

[KMjuniormint5]

ok but how do you start the calculations. . . .because you got too many unknowns

 

[Celunas]

Fwx and Fwy are just two projections of the Fw force, so you can write :

Fwx = Fw*cos(32°)
Fwy = Fw*sin(32°)

Then your equation has only one unknown, you can find it out and use your result to calculate the work. ( just be careful to use Fwx and not Fw in the work's calculation)

 

[KMjuniormint5]

but you don't know the force of the worker so you are unable to calculate Fw . ..

 

[Dick]

but you don't know the force of the worker so you are unable to calculate Fw . ..

If the block is moving at constant speed, then the horizontal component of the worker's push must cancel the frictional force. Otherwise there would be a net force and the box would accelerate. Leave the Fw as unknown. The balance of the horizontal and vertical forces will give you two equations in two unknowns and allow you to calculate Fw.

 

[Redbelly98]

Work with the last equation you got in post #1:

<br /> F_{wx} = \mu[(\text{Weight})+(F_{wy})] <br />

Celunas gave expressions for Fwx and Fwy, in terms of Fw. Substitute those expressions into the above equation. Then solve it for Fw. Everything else (mu and weight) are known.

 

[Celunas]

Indeed Redbelly98 that was my point.

With this expression you find Fw.

Then you can deduce Fwx ( Fwx = Fw * cos(32°) )

And once you have Fwx.. well, we won't give you the solution :)

 

[KMjuniormint5]

alright so I plugged everything and got Fw (force of the worker) to be 71.32 Newtons then Work is = to the dot product of the force multiplied by the distance. . .so I take 71.32*7.8 = 556.296 Joules . . .does that seem logical?

 

[Redbelly98]

alright so I plugged everything and got Fw (force of the worker) to be 71.32 Newtons then Work is = to the dot product of the force multiplied by the distance

Yes, you take the dot product. How do you take the dot product of two vectors? (Hint: it involves the angle between those vectors.)

. . .so I take 71.32*7.8 = 556.296 Joules . . .does that seem logical?

Not quite. This is the product, which is different than the dot product.

 

[KMjuniormint5]

(71.32*sin32)*7.8 = 294.79J
(71.32*cos32)*7.8 = 471.77 J

Add those together to get 766.56 J

 

[Redbelly98]

Looks like you need to review how to calculate dot products. Or it may be quicker to look up the formula for work, when the force and displacement are at an angle to one another. This formula is almost certainly in your textbook, in the chapter covering work.

 

[Celunas]

The dot product of the work of a force is :

vector F. vector d = F * d *cos (vector F,vector d)

In fact a force can work only in the direction of the movement (for a linear movement and a constant force). So only the part of the force being in the direction of the movement (Fwx for instance) works, because the whole movement lies in this direction.

Then, the work of the force is calculated by the product of the part of this force being in the movement's direction and the distance considered.

For any movement direction, any force F has a component on this direction F1 = F cos (angle between F and d).

This is why for any force F, the work of this force being the product of its component in the movement's direction and the distance is :
W = F*d*cos(vectorF,vectord), this is the definition of the dot product.

I hope I brought you something !