Work Done When Force Is Varying
Trending Questions
If the force and displacement of a particle in direction of force are doubled. Work would be
Double
times
Half
times
A force of acts on a body initially at rest. The work done by the force during the first second of motion of the body is
A particle moves under the effect of a force F = Cx from x = 0 to x=x1 . The work done in the process is
Cx21
12Cx21
Cx1
Zero
- +10 J
- +20 J
- −20 J
- −10 J
- 76 J
- 46 J
- 19 J
- 38 J
- −2ka2
- 2ka2
- −ka2
- ka2
A position dependent force F=7−2x+3x2 acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5m. The work done in joules is
70
270
35
135
- Work done against upthrust exerted by the water
- Work done against upthrust plus loss of gravitational potential energy of the block.
- Work done against upthrust minus loss of gravitational potential energy of the block.
- All of the above.
- ab(1−2e−bd)
- ab(2−e−bd)
- ab(1−e−bd)
- 2ab(2−e−bd)
- 16 J
- 6 J
- 0 J
- 12 J
F=⎧⎪⎨⎪⎩4 N0≤x≤66x N6<x≤103x2 N10<x≤20⎫⎪⎬⎪⎭ (where x is meters)
Find the total work done by this force to displaces from x=4 to x=15 metres.
- 0
- 2000 J
- 2500 J
- 2575 J
- 30 J
- 16 J
- 46 J
- 14 J
Given →F=(xy2)^i+(x2y)^jN. The work done by →F when a particle is taken along the semicircular path OAB where the coordinates of B are (4, 0) is
653J
752J
734J
Zero
The distance x moved by a body of mass 0.5 kg due to a force varies with time t as
x=3t2+4t+5
where x is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?
25 J
50 J
75 J
100 J
The distance x moved by a body of mass 0.5 kg due to a force varies with time t as
x=3t2+4t+5
where x is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?
25 J
50 J
75 J
100 J
- 15 J
- 12 J
- 32 J
- −15 J
- 4.5 J
- 13.5 J
- 9.0 J
- 18 J
- 12 J
- 9 J
- 6 J
- 3 J
A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x)=−kx+ax3. Here k and a are positive constants. For x≥0, the functional form of the potential energy U(x) of the particle is
- 12 J
- 9 J
- 6 J
- 3 J
- False
- True
- mg√2μ0
- mg2√2μ0
- mgμ0
- mg2μ0
F=⎧⎪⎨⎪⎩4 N0≤x≤66x N6<x≤103x2 N10<x≤20⎫⎪⎬⎪⎭ (where x is meters)
Find the total work done by this force to displaces from x=4 to x=15 metres.
- 0
- 2000 J
- 2500 J
- 2575 J
A force F=−K(y^i+x^j), where K is a positive constant, acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particle is
−2 Ka2
2 Ka2
−Ka2
Ka2