Work Done by Horse in 11.8 min: 190904 N*m

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The discussion focuses on calculating the work done by a horse pulling a cart with a force of 180.0 N at a 26.0-degree angle over 11.8 minutes. The calculations involve converting speed from kilometers per hour to meters per second and time from minutes to seconds, resulting in a distance of approximately 1180 meters. The formula used is W = Fcos(theta)d, but there is confusion regarding the correct time conversion and the final units of measurement. Ultimately, the correct answer is sought in kilojoules (kJ), clarifying the need for unit consistency in the calculations. The importance of precise time conversion and unit understanding is emphasized in the discussion.
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horse power

A horse pulls a cart with a force of 180.0 N at and angle of 26.0o with respect to the horizontal and moves along at a speed of 6.0 km/hr. How much work does the horse do in 11.8 min?

ok I use the equation Fcos(theta)d=W

6km/hr=1.67m/s
11.8min=708s

so 708*1.67=1182 m 1182m *180N*cos(26)=190904 N*m which is wrong
 
Last edited:
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Trousers said:
A horse pulls a cart with a force of 180.0 N at and angle of 26.0o with respect to the horizontal and moves along at a speed of 6.0 km/hr. How much work does the horse do in 11.8 min?

ok I use the equation Fcos(theta)d=W

6km/hr=1.67m/s
11.8min=720s

so 720*1.67=1180 m 1180m *180N*cos(26)=190904 N*m which is wrong

what I am doing wrong?

Power is watts = N*m/s

Edit: Oh I see they want work. I was confused by your title.
 
Last edited:
11.8 min is not 720 sec.
 
if I use 708 seconds it's still wrong
 
Trousers said:
if I use 708 seconds it's still wrong

I see that it still works out to 1180.

Perhaps they want the answer in Joules, or KJ
 
thanks it was in kJ
 
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