Question #77 from the Physical Sciences section:
A force F is used to raise a 4-kg mass M from the ground to a height of 5 m
What is the work done by the force F? (Note: sin 60o = 0.87; cos 60o = 0.50. Ignore friction and the weights of the pulleys.)
50 J
100 J
174 J
200 J
Answer: D
Explanation: Work is the product of force and distance. The easiest way to calculate the work in this pulley problem is to multiply the net force on the weight mg by the distance it is raised: 4 kg x 10 m/s2 x 5 m = 200 J. Therefore, answer choice D is the best answer.
W=Fdcos(theta)
I am confused at to when the "cos(theta)" part of the equation comes into play. If theta is the angle between the distance and the force applied, then why is the answer 200J and not 100J. Can I simply use W=Fd and ignore any mentions of angles when working with pulleys?
A force F is used to raise a 4-kg mass M from the ground to a height of 5 m
What is the work done by the force F? (Note: sin 60o = 0.87; cos 60o = 0.50. Ignore friction and the weights of the pulleys.)
50 J
100 J
174 J
200 J
Answer: D
Explanation: Work is the product of force and distance. The easiest way to calculate the work in this pulley problem is to multiply the net force on the weight mg by the distance it is raised: 4 kg x 10 m/s2 x 5 m = 200 J. Therefore, answer choice D is the best answer.
W=Fdcos(theta)
I am confused at to when the "cos(theta)" part of the equation comes into play. If theta is the angle between the distance and the force applied, then why is the answer 200J and not 100J. Can I simply use W=Fd and ignore any mentions of angles when working with pulleys?
