What is the Work Done by Several Forces?

In summary, the problem involves finding the magnitude of force and work done by the worker in moving a crate at constant velocity. There are five forces acting on the crate: the worker, gravity, normal force, friction, and an unknown force. The equations used are w = f * d * cos theta and w = mg. After solving for the acceleration, it is determined that the work done by the worker and friction is equal in magnitude, but opposite in sign due to the negative value of cosine at 180 degrees.
  • #1
Adrian Pete
6
0

Homework Statement


Problem # 1
t72Lb2d.png

Homework Equations


Work done by several forces

w=f * d cos theta
w=mg
n=w

The Attempt at a Solution


I can't figure out how to solve a) and b)
NOT SURE:
so my attempt a) f = mg = 30kg*9.8m/s^2 = 294N =magnitute of the force must the worker apply.(NOT SURE)
b) WofF= f d cos theta = 294N * 4.5m * cos0 = 1323N
?
 
Last edited:
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  • #2
Hi Adrian Pete and welcome to PF.

Your "relevant equations" show no equations. What do you mean by "work done by several forces"? How many forces do you have acting on the crate? How many of them are doing work?
 
  • #3
5 forces acting on it Sir
 
  • #4
Can you name these 5 forces and find how much work each one does?
 
  • #5
a) work must the worker apply = :(
b) work done by the force = :(
c) work done by friction = Wf = 73.5N*4.5m
d) work done by gravity = Wg = 294N*4.5m
e) work done by normal force Wn = 294N*4.5m
only these Sir :( can't figure out to solve a) and b)
 
  • #6
First, it seems you have misunderstood the questions (a) and (b). It also seems that question (a) has a typo. I believe the first two questions should read
(a) What magnitude of force (not work) must the worker apply?
(b) How much work is done by this force?
There are only 4 forces exerted by (1) the worker, (2) gravity, (3) normal force and (4) friction.
It also seems you think that Work = Force × Distance. This is not necessarily true and is not true for some of the forces here. Please look up the definition for work done by a constant force and apply it correctly.
Finally, you are told that the crate is moving at constant velocity. What must be true for this to be the case?
 
  • #7
force = m * a?
so a is 0 because it moves at constant velocity?

so force = 30kg * 0 m/s^2

a) f = 0N

so work is w = f d cos theta so,

w = 0 * 4.5m cos 0

so

b) w = 0J = work done by the worker?
 
  • #8
Adrian Pete said:
orce = m * a?
so a is 0 because it moves at constant velocity?
This is correct. It is not correct that the force exerted by the worker is zero. The worker must exert a force otherwise the crate will not move. Given that the crate is moving and that its acceleration is zero, what must be tru about the forces acting on the crate?
 
  • #9
since there is no acceleration therefore,

F - fk = ma = 0

F = fk ?

so F is basically fk?

so work done by F is the same as work done by fk?
 
  • #10
Adrian Pete said:
so work done by F is the same as work done by fk?
In magnitude. The work done by fk is negative while the work done by F is positive.
 
  • #11
oh okay, because cos 180 is -1. okay Sir. Thank you for the help ^ _ ^
 
  • #12
Adrian Pete said:
oh okay, because cos 180 is -1. okay Sir. Thank you for the help ^ _ ^
That's correct. You are welcome.
 

What is work done by several forces?

Work done by several forces is the energy transferred to an object when multiple forces act on it, causing it to move. It is a measure of the force applied over a distance.

How is the work done by several forces calculated?

The work done by several forces is calculated by multiplying the force applied by the distance over which it is applied. This can be represented by the equation W = F * d, where W is work, F is force, and d is distance.

What is the unit of measurement for work done by several forces?

The unit of measurement for work done by several forces is joules (J). This is the same unit used to measure energy.

Can the work done by several forces be negative?

Yes, the work done by several forces can be negative. This occurs when the force applied is in the opposite direction of the motion of the object. Negative work represents energy being taken away from the object.

How does the angle between the force and displacement affect the work done by several forces?

The angle between the force and displacement affects the work done by several forces through the use of the cosine function. The work done is calculated as W = F * d * cosθ, where θ is the angle between the force and displacement vectors. This means that if the force and displacement are in the same direction (θ = 0), the work done will be at its maximum. If they are perpendicular (θ = 90 degrees), the work done will be zero.

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