Q7. Truck driver load some oil drums into a truck by lifting them directly. Each drum has a mass of 60 kg and a platform of a truck is 0.6 m above the ground (g=10m/s). Force needed to lifta drum into the truck is ON O 600N O 0.06N O 6ON
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(i) F = mg = 80 x 10 = 800N.
(ii) Energy = m x g x h = 80 x 10 x 0.8 = 640J.
(iii) In this case F = mg sinθ = 800 x 0.8/3 = 640/3 = 213.3N
It is less than the answer in part (i) as the efforts acts through 3 m whereas the weight is lifted up through 0.8 m. The effort needed may be 300 N instead of 213.3 N owing to loss of energy in overcoming friction and air resistance.
(iv) The energy released owing to combustion of fuel changes into kinetic energy of the truck and load, and this changes in heat energy of the pebbles on the road and of wheels.
(v) The kinetic energy is dissipated as heat.
Explanation:
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