An object with a mass of $12$ $12$ $k g$ $kg$ is on a surface with a kinetic friction coefficient of $1$ $1$ . How much force is necessary to accelerate the object horizontally at $7$ $7$ $m s^{- 2}$ $ms^-2$ ?

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An object with a mass of $12$ $12$ $k g$ $kg$ is on a surface with a kinetic friction coefficient of $1$ $1$ . How much force is necessary to accelerate the object horizontally at $7$ $7$ $m s^{- 2}$ $ms^-2$ ?

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Answer 1 · 2016-02-13T23:52:14

The total force is composed of two pieces: the force required to accelerate the mass and the force required to overcome friction. $F = m a + μ m g = 12 \cdot 7 + 1 \cdot 12 \cdot 9.8 = 201.6$ $F=ma+mumg=12*7+1*12*9.8=201.6$ $N$ $N$ .

Explanation:

The total force is made up of two different forces.

If there were no friction, the force required to accelerate a mass of $12$ $12$ $k g$ $kg$ at an acceleration of $7$ $7$ $m s^{- 2}$ $ms^-2$ is given by Newton's Second Law :

$F = m a = 12 \cdot 7 = 84$ $F=ma=12*7=84$ $N$ $N$

The second part is the frictional force:

$F_{f} = μ F_{N}$ $F_f=muF_N$ where $μ$ $mu$ is the frictional coefficient and $F_{N}$ $F_N$ is the normal force, which in turn is $m g$ $mg$ , the mass times the acceleration due to gravity.

$F_{f} = μ m g = 1 \cdot 12 \cdot 9.8 = 117.6$ $F_f=mumg=1*12*9.8=117.6$ $N$ $N$

To find the total force, just add these two forces:

$F = m a + μ m g = 12 \cdot 7 + 1 \cdot 12 \cdot 9.8 = 201.6$ $F=ma+mumg=12*7+1*12*9.8=201.6$ $N$ $N$ .

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An object with a mass of $12$ $12$ $k g$ $kg$ is on a surface with a kinetic friction coefficient of $1$ $1$ . How much force is necessary to accelerate the object horizontally at $7$ $7$ $m s^{- 2}$ $ms^-2$ ?

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An object with a mass of $12$ $12$ $k g$ $kg$ is on a surface with a kinetic friction coefficient of $1$ $1$ . How much force is necessary to accelerate the object horizontally at $7$ $7$ $m s^{- 2}$ $ms^-2$ ?