A 2 kg box is sliding across ice at 2 m/s. If 40 J of work is done on the box ...?
2019-02-04 1:50 pm
A 2 kg box is sliding across ice at 2 m/s. If 40 J of work is done on the box in the same direction that it is moving (i.e. theta=0), how fast is the box sliding now? Round to one decimal. m/s
回答 (2)
2019-02-04 2:06 pm
Increase in kinetic energy = Work done on the box
(1/2)mv² - (1/2)mvₒ² = 40
(1/2) × 2 × v² - (1/2) × 2 × 2² = 40
v² - 4 = 40
v² = 44
Final velocity, v = √44 = 6.6 m/s
(1/2)mv² - (1/2)mvₒ² = 40
(1/2) × 2 × v² - (1/2) × 2 × 2² = 40
v² - 4 = 40
v² = 44
Final velocity, v = √44 = 6.6 m/s
2019-02-04 2:26 pm
Ef = Eo+E = 2/2*2^2+40 = 44 joule
Vf = √2Ef/m = √44 = 2√11 m/sec ( ≈ 6.633 )
Vf = √2Ef/m = √44 = 2√11 m/sec ( ≈ 6.633 )
收錄日期: 2021-05-02 11:13:58
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