Question

A team of dogs drags a 53.9 kg sled 1.62 km over a horizontal surface at a constant speed. The coefficient of friction between the sled and the snow is 0.234. The acceleration of gravity is 9.8 m/s 2 . Find the work done by the dogs. Answer in units of kJ.

Answer 1

Answer:

200.24 kJ

Explanation:

$m$ = mass of sled = 53.9 kg

$d$ = distance traveled by the sled = 1.62 km = 1620 m

$\mu$ = Coefficient of friction between sled and snow = 0.234

frictional force acting on the sled is given as

$f = \mu mg$

$F$ = Applied force by the dogs on the sled

Since the sled moves at constant speed, the force equation for the motion of the sled is given as

$F = f \\F = \mu mg$

$W$ = Work done by the dogs on the sled

Work done by the dogs on the sled is given as

$W = F d\\W = \mu mg d\\W = (0.234) (53.9) (9.8) (1620)\\W = 200237.64 J\\W = 200.24 kJ$

Answer 2

The work done by the dogs is 200.238 kJ.

The work done by the dog is equal to the work done to move the sled through the distance and the work done against friction.

Formula:

• W = dma+μmgd............. Equation 1

Where:

• W = work done by the dogs
• m = mass of the sled
• a = acceleration of the sled
• g = acceleration due to gravity
• μ = coefficient of friction

From the question,

Given:

• m = 53.9 kg
• a = 0 m/s² (move with constant speed)
• d = 1.62 km = 1620 m
• g = 9.8 m/s²
• μ = 0.234

Substitute these values into equation 1

• W = (53.9×0×1620)+(53.9×9.8×0.234×1620)
• W = 2002378 J
• W = 200.238 kJ

Hence, The work done by the dogs is 200.238 kJ.

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