Finding work done by downward force and increase in thermal energy

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To determine the work done by the worker's force and the increase in thermal energy of the block-floor system, the worker's force must be analyzed in both horizontal and vertical components. The worker is pushing a 23 kg block at a constant speed, indicating that the net force is zero. The equations derived from the horizontal and vertical components of the forces allow for the calculation of the normal force and the work done against friction. Given the coefficient of kinetic friction of 0.40, the problem can be solved using the established relationships. The conditions provided ensure a unique solution for the worker's force exerted at the specified angle.
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A worker pushed a 23 kg block 13 m along a level floor at constant speed with a force directed 22° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.40, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?


Fh=Fcos(-22)
Fv=Fsin(-22)


To calculate the normal force I need both the gravitational force and the vertical component of the worker's force, and for horizontal work I also obviously need the horizontal component of their force. So how do I find either of those if I'm not given the magnitude of the worker's force?
 
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rockchalk1312 said:
A worker pushed a 23 kg block 13 m along a level floor at constant speed with a force directed 22° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.40, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?Fh=Fcos(-22)
Fv=Fsin(-22)To calculate the normal force I need both the gravitational force and the vertical component of the worker's force, and for horizontal work I also obviously need the horizontal component of their force. So how do I find either of those if I'm not given the magnitude of the worker's force?

The key is that the worker is pushing the block at constant speed, so acceleration (and therefore net force) must be zero.

So, for the horizontal components we have:

Fw cos(22) - Fn(0.40) = 0 where Fw is the force from the worker and Fn the normal force on the block.

For the vertical components we have:

Fw sin(-22) - mg + Fn = 0.

Two equations and two unknowns. You know what to do.

A way to check--before attempting--if a problem like this is solvable is think if the conditions given to you uniquely determine the problem. That is, for a block of a given mass and a force of a given angle and a given coefficient of friction, are there more than one possible forces that the worker could exert on the block at the given angle for it to move at a constant velocity? Of course not!
 
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