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Physics question. A boy pulls a 26.0 kg box...?
A boy pulls a 26.0 kg box with a 140 N force at 32° above a horizontal surface. If the coefficient of kinetic friction between the box and the horizontal surface is is 0.21 and the box is pulled a distance of 22.0 m, what is the work done by the friction force on the box?
3 Answers
- BrainardLv 73 years agoFavorite Answer
Let R = normal reaction and F = frictional force
Then
R + 140sin(32) = 26g,
R = 26g - 140sin(32)
= 180.6111303
= 180.61 N
F = 0.21 x 180.61
= 37.93 N
Work done by friction = 37.93 x 22
= 834.46 J
- electron1Lv 73 years ago
There are two forces that affect the friction force. They are the weight of the box, and the vertical component of the force.
Weight = 26.0 * 9.8 = 254.8 N
N = 254.8 – 140 * sin 32
Ff = μ * N = 53.508 – 29.4 * sin 32
This is approximately 37.9. N
Work = Ff * d = (53.508 – 29.4 * sin 32) * 22
Work = 1177.176 – 646.8 * sin 32
This is approximately 834 N * m. I hope this is helpful for you.
- RealProLv 73 years ago
Normal force = mg - Fsin32° = 180 N
Force of friction = 180 N * 0.21 = 37.9 N
Work of friction = force of friction * displacement = 830 J
