A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in figure. If it starts its journey from rest at x = 0, Work Done from x=0 to x = 12 m is:
Area bounded by force time graph represents work done so,
Work= ½(12+4)×10= 80 J
A chord is used to lower vertically a block of mass M a distance d at a constant downward acceleration of g/4. Then the work done by the cord on the block is :
Forces acting on the object are , mg in downward direction and Tension force T in upward direction, as the object moves downward.
mg - T= ma= mg/4
so, T= 3mg/4
Work done by tension force is W= T.d.cos (180°)= -3mgd/4
An object of mass m is tied to string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases (it is pulled extremely slowly so that equilibrium exists at all times) until the string makes an angle θ with the vertical. Work done by the force F is :
Be definition of Work, Work Done = F.S. cos θ
Taking S and cos θ together like
W= F. (S cos θ)= Force × Component of displacement along force
Here Force F is acting horizontally and displacement toward right is L sin θ, So
Work done = F.L. sin θ
An object of mass m is tied to string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases (it is pulled extremely slowly so that equilibrium exists at all times) until the string makes an angle θ with the vertical. Work done by the gravitational force mg is :
Be definition of Work, Work Done = F.S. cos θ
Taking F and cos θ together like
W= (F. cos θ) . s = Component of Force along displacement × displacement
Here gravitational force mg is acting downwards and displacement in upward direction is (L- L cos θ), So
Work done = mg.(L-Lcosθ)cos (180°)= -mgL(1-cosθ)
A force of (3ˆi + 4ˆj)N acts on a body and displaced it by (3ˆi + 4ˆj)m . The work done by the force is
By Definition of Work ,
Work Done = F.S= (3ˆi + 4ˆj).(3ˆi + 4ˆj)= 9+16 =25 J
When the bob of a simple pendulum swings, the work done by tension in the string is
As Angle between tension force and displacement is 90°. Work done is zero
A particle moves along the x-axis from x = 0 to x = 5 m under the influence of a force F (in N) given by F = 3x² –2x + 7. Calculate the work done by this force.
Work Done by Variable force is ∫F.dx, So
W= ∫(3x² –2x + 7).dx=[3x³/3-2x²/2+7x]= [x³-x²+7x]
Using Proper limits from x=0 to x= 5, we get
W= 125-25+35=135 J
A force of (5 + 3x) N, acting on a body of mass 20 kg along the x-axis, displaces it from x = 2 m to x = 6 m. The work done by the force is
Work Done by Variable force is ∫F.dx, So
W= ∫(5 + 3x).dx= 5x+3x²/2
Using Limits from x=2 to x=6 we get
W= 68 J
A person draws water from a 5m deep well in a bucket of mass 2 kg of capacity 8 litre by a rope of mass 1 kg. What is the total work done by the person ? [g = 10 m/s2]
Work Done in Pulling the Bucket + Rope System=
Mgh+mg(h/2)
10×10× 5 + 1×10×(5/2) =525 J