Find the force required to start a 1 ton object moving along a surface with a coefficient of friction of 0.62.

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Multiple Choice

Find the force required to start a 1 ton object moving along a surface with a coefficient of friction of 0.62.

Explanation:
To determine the force required to start a 1-ton object moving along a surface with a coefficient of friction of 0.62, you need to apply the concept of frictional force. The force of friction can be calculated using the formula: \[ F_{friction} = \mu \times F_{normal} \] Where: - \( F_{friction} \) is the force of friction, - \( \mu \) is the coefficient of friction, - \( F_{normal} \) is the normal force, which, on a horizontal surface, is equal to the weight of the object. First, convert the weight of the object from tons to kilograms. Since 1 ton is equivalent to approximately 1000 kg, we have: \[ Weight = 1 \, \text{ton} = 1000 \, \text{kg} \] Next, calculate the normal force, which is the weight of the object in Newtons. The force due to gravity (weight) is given by: \[ F_{normal} = m \times g \] Where: - \( m \) is the mass (1000 kg), - \( g \) is the acceleration due to gravity (approximately 9

To determine the force required to start a 1-ton object moving along a surface with a coefficient of friction of 0.62, you need to apply the concept of frictional force. The force of friction can be calculated using the formula:

[ F_{friction} = \mu \times F_{normal} ]

Where:

  • ( F_{friction} ) is the force of friction,

  • ( \mu ) is the coefficient of friction,

  • ( F_{normal} ) is the normal force, which, on a horizontal surface, is equal to the weight of the object.

First, convert the weight of the object from tons to kilograms. Since 1 ton is equivalent to approximately 1000 kg, we have:

[ Weight = 1 , \text{ton} = 1000 , \text{kg} ]

Next, calculate the normal force, which is the weight of the object in Newtons. The force due to gravity (weight) is given by:

[ F_{normal} = m \times g ]

Where:

  • ( m ) is the mass (1000 kg),

  • ( g ) is the acceleration due to gravity (approximately 9

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