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Difficulty: MediumWork, Energy and Power

An electric engine delivers a constant power of 600 W600\text{ W} to move a 50 kg50\text{ kg} box at a constant speed of 4 m s−14\text{ m s}^{-1} along a horizontal surface. What is the coefficient of kinetic friction between the box and the surface? (Take g=10 m s−2g = 10\text{ m s}^{-2})

  1. 0.300.30Answer
  2. B
    1.201.20
  3. C
    3.003.00
  4. D
    4.804.80

Answer

The coefficient of kinetic friction between the box and the surface is 0.30.
The option stating 0.30 is correct because the power delivered by the engine P=FvP = F v yields an applied force of 150 N150\text{ N}. Since velocity is constant, the frictional force equals 150 N150\text{ N}. Dividing by the normal force N=mg=500 NN = mg = 500\text{ N} gives μk=0.30\mu_k = 0.30.

Step-by-Step Solution

1
Calculate the pulling force exerted by the engine using the relationship between power, force, and velocity.
F=Pv=600 W4 m s−1=150 NF = \frac{P}{v} = \frac{600\text{ W}}{4\text{ m s}^{-1}} = 150\text{ N}
Power is defined as work per unit time, or force multiplied by constant speed.
2
Determine the frictional force acting on the box.
fk=F=150 Nf_k = F = 150\text{ N}
Since the box moves at a constant velocity, net horizontal force is zero, meaning pulling force equals kinetic friction force.
3
Calculate the normal reaction force exerted by the surface on the box.
N=mg=50 kg×10 m s−2=500 NN = m g = 50\text{ kg} \times 10\text{ m s}^{-2} = 500\text{ N}
On a horizontal plane, normal reaction equals the weight of the object.
4
Compute the coefficient of kinetic friction.
μk=fkN=150 N500 N=0.30\mu_k = \frac{f_k}{N} = \frac{150\text{ N}}{500\text{ N}} = 0.30
The coefficient of kinetic friction is the ratio of kinetic friction force to normal reaction force.

Key Concept

Power, Work against Friction, and Coefficient of Kinetic Friction
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Topics

Work, Energy and Power

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