Work Done, Varying Force and Path
Trending Questions
- 9 Ns
- Zero
- 0.9 Ns
- 1.8 Ns
A body of mass 3 kg is under a force, which causes a displacement in it is given by S=t33 (in m). Find the work done by the force in first 2 seconds
- 2 J
- 3.8 J
- 5.2 J
- 24 J
Consider π3=1 for calculation and take g=10 m/s2)
- 10√3 m/s
- 10 m/s
- 5√3 m/s
- 3√3 m/s
- 400 J
- −400 J
- −200 J
- 200 J
- 1
- 10
- 50
- 100
- d=l4
- d=l2
- d=l√2
- d=l
- the weight reading at A is greater than the weight reading at E by 2W.
- the weight reading at G is W.
- the ratio of the weight reading at E to that at A is 0.
- the ratio of the weight reading at A to that at C is 2.
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). Find the total work done by the forces →F on the particle.
Ka2
-2 Ka2
2 ka2
- W2=W1
- W2=2W1
- W2=3W1
- W2=4W1
- −4ka2
- 2ka2
- −2ka2
- 4ka2
- 6.25 J
- 12.50 J
- 18.75 J
- 25.00 J
[Take, g=10 m/s2]
- 3.0 m/s
- 4.0 m/s
- 8.2 m/s
- 3.96 m/s
A person is holding a bucket by applying a force of 10 N. He moves a horizontal distance of 5 m and then climbs up a vertical distance of 10 m. The total work done by him is equal to
- 50 J
- 150 J
- 100 J
- 200 J
- −2Ka2
- 2Ka2
- −Ka2
- Ka2
- 275 J
- 250 J
- 475 J
- 450 J
- d=l4
- d=l√2
- d=l2
- d=l
- Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
- Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
- Assertion is correct but Reason is incorrect
- Both Assertion and Reason are incorrect
- 0.9 Ns
- 1.8 Ns
- 9 Ns
- Zero
- 3k(θ0)2ℓ
- 2k(θ0)2ℓ
- k(θ0)2ℓ
- k(θ0)22ℓ
- 10 J
- 20 J
- 40 J
- 30 J
- √2×10−2 Ns
- √3×10−2 Ns
- √5×10−2 Ns
- 2×10−2 Ns
(all quantities are in SI units)
- 1
- 32
- 2
- 12
- 24J
- 96J
- 48J
- 6J
- 0.5 joule
- 0.1 joule
- 0.2 joule
- 0.3 joule
- 9 Ns
- Zero
- 0.9 Ns
- 1.8 Ns
A force of 50 dynes is acted on a body of mass 5 g which is at rest for an interval of 3 seconds, then impulse is
0.15×10−3N−s
0.98×10−3N−s
1.5×10−3N−s
2.5×10−3N−s
- 6.25 J
- 12.50 J
- 18.75 J
- 25.00 J
- l=aA−bB√A2+B2
- l=ab+AB√A2+B2
- l=ab−AB√A2−B2
- None of these
- 0
- 1.8 N−s
- 9 N−s
- 0.9 N−s