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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?

Respuesta :

Answer:

Explanation:

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

onyebuchinnaji

(a) The expression for the work done by the pulling force is [tex]\frac{\mu_k mg \times d}{cos\theta + \mu_k sin\theta}[/tex]

(b) The work done by the pulling force is 1652.5 J.

(c)The expression for the work done by the frictional force is [tex]- \frac{\mu_k mg \times d}{cos\theta + \mu_k sin\theta}[/tex]

(d) The work done by the frictional force is 1652.5 J.

Let the pulling force = F

The normal force on the sled is calculated as follows;

[tex]F_n = mg - Fsin(\theta)\\\\[/tex]

The frictional force between the sled and surface;

[tex]F_k = \mu_k F_n\\\\F_k = \mu_k (mg - Fsin(\theta))[/tex]

The net horizontal force on the sled is calculated as follows;

[tex]Fcos(\theta) - F_k = 0\\\\Fcos(\theta) = F_k \\\\Fcos(\theta) = \mu_k (mg - Fsin\theta)\\\\Fcos\theta = \mu_k mg - \mu_k Fsin\theta \\\\Fcos\theta + \mu_k Fsin\theta = \mu_k mg\\\\F( cos\theta + \mu sin\theta) = \mu_k mg\\\\F = \frac{\mu_k mg}{cos\theta + \mu sin\theta} \\\\Fd = \frac{\mu_k mg}{cos\theta + \mu sin\theta} \times d \\\\W = \frac{\mu_k mg\times d}{cos\theta + \mu sin\theta}[/tex]

The expression for the work done by the pulling force is [tex]\frac{\mu_k mg \times d}{cos\theta + \mu_k sin\theta}[/tex]

(b)

the work done by the pulling force is calculated as follows;

[tex]W = \frac{\mu_ k mg\times d}{cos \theta + \mu_ k sin\theta} \\\\W = \frac{0.13 \times 52.3 \times 9.8 \times 21.8 }{cos(36.7) \ + \ 0.13sin\times (36.7)} \\\\W = 1652.5 \ J[/tex]

(c)

the work done by friction is calculated as follows;

[tex]Fcos(\theta) \times d = F_k\times d \\\\Fcos(\theta) = \mu_k (mg - Fsin(\theta)) d \\\\Fcos(\theta)d = \mu_k mgd - \mu_k F sin(\theta) d \\\\Fcos(\theta)d+ \mu_k F sin(\theta) d = \mu_k mg d \\\\Fd (cos(\theta) + \mu_k sin(\theta)) = \mu_k mg d\\\\Fd = \frac{\mu_k mg \times d}{cos(\theta) + \mu_k sin(\theta)}\\\\[/tex]

[tex]-W_k = \frac{\mu_k mg \times d}{cos(\theta) + \mu_k sin(\theta)}[/tex]

(d)

Since the sled moves at a constant speed, the magnitude of work done by friction is equal to work done by the applied force;

[tex]-W_k = \frac{\mu_k mg \times d}{cos(\theta) + \mu_k sin(\theta)} = W = 1652.5 \ J\\\\W_k = - 1652.5 \ J[/tex]

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