Work Energy Theorem
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In physics What is the sign convention used for work done, in thermodynamics.
Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take 'g' constant with a value 10 m/s2. The work done by the (i) gravitational force (ii) resistive force of air is
- (i) 1.25 J, (ii) –8.25 J
- (i) 100 J, (ii) 8.75 J
- (i) 10 J, (ii) –8.75 J
- (i) –10 J, (ii) –8.25 J
- 1 m/s
- 2 m/s
- 3 m/s
- 4 m/s
- 30 J
- 40 J
- 50 J
- 20 J
- 12mv2t21t2
- mv2t21t2
- 2mv2t21t2
- 12(mt1)2t2
- GmM12R
- GmM6R
- GmM3R
- GmMR
- 5 m
- 1 m
- 2 m
- 4 m
- 2.5 cm
- 5.5 cm
- 11.0 cm
- 8.5 cm
A block of mass m lying on a smooth horizontal surface is attached to a spring (of negligible mass) of spring constant k. The other end of the spring is fixed as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is
- F√mk
- πF√mk
- 2F√mk
- Fπ√mk
- 1 m/s
- 2 m/s
- 2√2 m/s
- 4 m/s
- 20 m/s
- 10√3 m/s
- 5√2 m/s
- 10 m/s
- √2(u2−gl)
- √u2−gl
- u−√u2−2gl
- √2gl
If force , work and velocity are taken as fundamental quantities. What is the dimensional formula of time?
- logex
- x
- ex
- x2
- 61 J
- 32 J
- 102 J
- 46 J
[Take g=10 m/s2, √1552−40√3≈38.50 & 10+2√3≈13.5 for calculation. Neglect the effect of corner]
- 2.5 m
- 3 m
- 2.1 m
- 2 m
- 12000 J
- −12000 J
- −4500 J
- −9300 J
- Zero
- 1.5x
- 2x
- 2.5x
- 2/3 kg m/s
- −3/2 kg m/s
- 3/2 kg m/s
- −2/3 kg m/s
- MV2T
- 12MV2T2
- MV2T2
- 12MV2T
An object of mass is moving with a constant velocity, . How much work should be done on the object in order to bring the object to rest?
- 0.24
- 0.36
- 0.12
- 0.48
The momentum of a body increases by 20%. Calculate the percentage increase in its K.E.
- h=v202g(1+fW)
- h=v202g(1−fW)
- h=v203g(1+fW)
- h=v202g(1+2fW)
- 2 MgLn2
- MgLn2
- MgL2n2
- nMgL
- 975 J
- 1025 J
- 725 J
- 825 J
- speed
- momentum
- energy
- velocity