TBR CBT 3 #49
For the pulley system shown below, how much energy is needed to lift a box weighing 4 kg to a height of 5 meters? (Note: cos 60˚ = 0.50, sin 60˚ = 0.87.)
C.) is the answer.
why can't i use Work = Fd cos(theta) to solve this? read the solution & understand it. but i ask because that is the equation that first popped up in my head so i went with it, but ended up with a wrong answer.
For the pulley system shown below, how much energy is needed to lift a box weighing 4 kg to a height of 5 meters? (Note: cos 60˚ = 0.50, sin 60˚ = 0.87.)
C.) is the answer.
why can't i use Work = Fd cos(theta) to solve this? read the solution & understand it. but i ask because that is the equation that first popped up in my head so i went with it, but ended up with a wrong answer.
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