Work done by a variable force problems
Hello candidate,
The work done by a force is calculated as the force applied multiplied by the distance travelled by the body by that applied force in the same direction or at a specific angle.
The work done by a variable force depends as a dependence of force on the displacement travelled which is basically solved using the method of integration from the initial point to the final point of displacement.
Hope you found it helpful. If you have any further queries feel free to post it here!!
dy/dx=sin ( x + y) + cos ( x + y )
Hello Aspirant,
Hope you are doing well!!
It is quite difficult to make integration sign, In place of integration sign I'll put |, and in place of power I'll put ** when you are solve in your notebook you can place | sign to integration sign and ** to power sign.
Let X+Y = v
1 + dY/dX = dv/dX
dY/dX= dv/dX-1
dY/dX = sin (X + Y) + cos (X + Y)
dv/dX - 1 = sin v + cos v
dv / (1+cos v + sin v) = dX
Integrate both side
| dv / (1 + cos v + sin v ) = | dX | dv / ( 1 + ((1 - tan**2 (v/2)) / (1 + tan** 2 (v/2)) + ((2 tan (v/2) / (1 + tan **2 (v/2))
= |dX | sec**2 (v/2) dv / (2(1+ tan(v/2))
=|dx log (1 + tan (v/2)
= X+ c log ( 1+ tan (X + Y) / 2) = X + c
I hope this will help you.
Feel free to ask any query.
the terms work and power . How will you evaluate the work done by a variable force?
Dear Student,
- The work done by a constant force magnitude F on a point that moves a displacement d in the direction of the force is the product of , W = Fd.
- Integration method can be utilized to calculate work done by a variable force and work done by a constant force
Explain the terms work and power . How will you evaluate the work done by a variable force? 22. State the law of conservation of energy .Illustrate this law in the case of a freely falling body.
Dear aspirant,
Work is the energy transferred from the object via the application of force along the displacement.
Power is basically the amount of energy transferred per unit time.
Work done by a variable force is the integral of F.dx where F is the force and dx is the displacement.
Explain the reason for the operation of the law of retun to a variable factor
Hi.
This law exhibits the short-run production functions in which one factor varies while the others are fixed.
Also, when you obtain extra output on applying an extra unit of the input, then this output is either equal to or less than the output that you obtain from the previous unit.
The Law of Variable Proportions concerns itself with the way the output changes when you increase the number of units of a variable factor. Hence, it refers to the effect of the changing factor-ratio on the output.
linear equation in one variable
Hello student,
The linear equation in one variable is an equation which can be expressed in the form such as 'ax + b = 0' (a, b and c are real numbers) where a and b are the two integers , and x has just one solution also which is a variable .
Taking example of one such equation :
4x+9=-11
When you solve this equation, it will have only one solution.
These type of equations are also known as first degree equations because the exponent on the variable is 1.
I hope the answer is clear to you. Thank you.
Given production is 1,00,000 units, fixed costs is Rs 2,00,000 Selling price is Rs 10 per unit and variable cost is Rs 6 per unit. Determine profit using technique of marginal costing
Hello aspirant,
Here is the answer to your question.
Marginal costing shows that the meaning of profit gets changed, but here profit means contribution.
Hence, Selling Price= 10 per unit-Variable Cost=6 per unit contribution which is equal to 4 per unit total contribution= 4 per unit * 100000 units= 4 lakhs
So, final profit= 4 lakhs - 2 lakhs (Fixed Cost= 2 lakhs.
Hope, it helps you.
the locus of the foot of the perpendicular draw from origin to a variable line passing through fixed point (2,3) is a circle whose diameter is
Hello aspirant,
The answer will be-
Assume slope of line be m
(2,3) are fixed points
Equation of line:-
y-3=m*(x-2)
y-3=mx-2m
mx-y+3-2m=0
Foot of perpendicular from origin (0,0) is-
(2-0)/m=(y-0)/(-1)= -(0+3-2m)/{0.5^(1+m^2)}
m= -x/y
Now, mx-y+3-2m=0
(-x/y)x-y+3-2(-x/y)=0
x^2+y^2-2^x-3^y=0
Centre=(1, 3/2)
Radius= 0.5^(1^2+(3/2)^2-0) = √13/2
Diameter = 2*Radius = 2* (√13/2) = √13
Hope, it helps you.