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 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
Two masses of 1 g and 4 g are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is
Kinetic Energy is given by K = ½mv² and Momentum is P=m.v, relation between kinetic energy and momentum is
K= P²/2m so P=√(2mK)
So P1/P2= √(m1)/√(m2)= √1/√4 = 1:2
If the kinetic energy of a body is increased by 300%, its momentum will increase by :
Relation between Kinetic Energy and Momentum is
P = √(2mK)
When kinetic energy is increased by 300 % it becomes 4K so new momentum is
P1= √(2m.4K) =2P
If the kinetic energy of a body is increased by 3%, its momentum will increase by :
A man pulls a bucket full of water from h metre deep well. If the mass of rope is m and mass of bucket full of water is M, then work done by the man is
Bucket of mass M will rise by a distance of h.
While center of mass of the rope rises by height of h/2.
So, Work Done = (M+m/2).g.h