A sled plus passenger with total mass m = 52.3 kg is pulled a distance d = 21.8 m across a horizontal, snow-packed surface for which the coefficient of kinetic friction with the sled is μk = 0.13. The pulling force is constant and makes an angle of φ = 36.7 degrees above horizontal. The sled moves at constant velocity. a) Write an expression for the work done by the pulling force in terms of m, g (acceleration due to gravity), φ, μk, and d. b) What is the work done by the pulling force, in joules? c) Write an expression for the work done on the sled by friction in terms of m, g (acceleration due to gravity), φ, μk, and d. d) What is the work done on the sled by friction, in joules?

Question

Physics

Brenden

23 April, 09:17

A sled plus passenger with total mass m = 52.3 kg is pulled a distance d = 21.8 m across a horizontal, snow-packed surface for which the coefficient of kinetic friction with the sled is μk = 0.13. The pulling force is constant and makes an angle of φ = 36.7 degrees above horizontal. The sled moves at constant velocity. a) Write an expression for the work done by the pulling force in terms of m, g (acceleration due to gravity), φ, μk, and d. b) What is the work done by the pulling force, in joules? c) Write an expression for the work done on the sled by friction in terms of m, g (acceleration due to gravity), φ, μk, and d. d) What is the work done on the sled by friction, in joules?

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Answers (1)

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Lovely · Answer 1

Since the sled plus passenger moves with constant velocity, force applied will be equal to frictional force. Let the force applied be F

a) Frictional force = μ R = F cosφ

R = mg - F sinφ

μ (mg - F sinφ) = F cosφ

μmg = F (μsinφ+cosφ)

F = μmg / (μsinφ+cosφ)

Work done

= F cosφ x d

= μmg x cosφ x d / (μsinφ+cosφ)

b) Work done

= 0.13 x 52.3 x 9.8 cos36.7 x 21.8 / (0.13 sin36.7 + cos36.7)

= 1164.61 /.87946

1324.23 J

c) work done on the sled by friction

= - (work done by force)

= - μmg x cosφ x d / (μsinφ+cosφ)

d) work done on the sled by friction

= - 1324.23 J