Solving Simple Work Problem: 515J

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SUMMARY

The sailor's work done while climbing a frictionless rope ladder is calculated using the formula W = Fd, where F is the force component acting along the direction of motion. The sailor's weight is 70.0 kg, and the ladder's angle with the mast is 30.0°. The correct calculation involves using the sine of the angle with respect to the vertical, leading to the conclusion that the work done is not 515 J as initially calculated. The correct approach requires adjusting the angle used in the sine function to reflect the angle with respect to the vertical rather than the horizontal.

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Mr Davis 97
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Homework Statement


A 70.0 kg sailor climbs an 11.5 m long rope ladder to a mast above at constant velocity. The rope ladder is at an angle of 30.0° with the mast. Assume that the ladder is frictionless. How much work does the sailor do?

Homework Equations


##W = Fd##

The Attempt at a Solution


[/B]
##F_{net||} = F_{a} + F_{g||}##
##F_{a} = -F_{g} = -mg\sin \theta##
##W = F_{a}d##
##W = -mgd\sin \theta##
##W = -(70kg)(-9.8m/s^{2})(11.5m)0.5##
##W = 515 J##

However, in the solution manual, this is not the correct answer. What am I doing wrong?
 
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Mr Davis 97 said:

Homework Statement


A 70.0 kg sailor climbs an 11.5 m long rope ladder to a mast above at constant velocity. The rope ladder is at an angle of 30.0° with the mast. Assume that the ladder is frictionless. How much work does the sailor do?

Homework Equations


##W = Fd##

The Attempt at a Solution


[/B]
##F_{net||} = F_{a} + F_{g||}##
##F_{a} = -F_{g} = -mg\sin \theta##
##W = F_{a}d##
##W = -mgd\sin \theta##
##W = -(70kg)(-9.8m/s^{2})(11.5m)0.5##
##W = 515 J##

However, in the solution manual, this is not the correct answer. What am I doing wrong?
The angle, θ, that you used is with respect to the deck, so it's not 30° .

That is to say, if you use sin(θ) that is with respect to the deck.
 
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