You pull a 50 kg sled along the ground. The coefficient of kinetic friction betw

Question

You pull a 50 kg sled along the ground. The coefficient of kinetic friction between the sled and the ground is 0.40. How much work do you do on the sled if you:

a) Pull the sled with a horizontal force, so that the sled moves at a constant speed, over a distance of 5 meters?

b) Pull the sled with a force at a 45 degree angle, so that the sled moves at a constant speed, over a distance of 5 meters?

c) Let’s call the motion in part (a) “the sled pull.” How many sled pulls would you have to do to burn 100 calories? How many would you have to do to burn the equivalent energy in a gallon of gasoline?

Explanation / Answer

a) Work = F.X = umg*X = 0.40*50*9.8*5 = 980 J ( F = umg)

b) Work = F.X = umg/(cos(45)+sin(45) )*X = 0.40*50*9.8*5*0.707 = 692.86 J

As ( Fcos(45) - (umg-Fsin(45) )= 0, => F = umg/(cos(45)+sin(45) )

c) 100 calories = 418.4 Joule

so effectively 2.34 sled can be pulled in both cases

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