An object with a mass of $5$ $5$ $k g$ $kg$ is on a surface with a kinetic friction coefficient of $4$ $4$ . How much force is necessary to accelerate the object horizontally at $6$ $6$ $m s^{- 2}$ $ms^-2$ ?

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An object with a mass of $5$ $5$ $k g$ $kg$ is on a surface with a kinetic friction coefficient of $4$ $4$ . How much force is necessary to accelerate the object horizontally at $6$ $6$ $m s^{- 2}$ $ms^-2$ ?

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Answer 1 · 2016-05-07T10:45:24

$F = F_{frict} + F_{accel} = 196 + 30 = 226$ $F=F_"frict"+F_"accel"=196+30=226$ $N$ $N$

Explanation:

The total force required will be given by $F = F_{frict} + F_{accel}$ $F=F_"frict"+F_"accel"$ - the sum of the force required to overcome friction and the force required to accelerate the mass.

The frictional force is given by:

$F_{frict} = μ F_{norm}$ $F_"frict"=muF_"norm"$ where the normal force $F_{norm} = m g$ $F_"norm"=mg$

So the friction force is $F_{frict} = μ m g = 4 \times 5 \times 9.8 = 196$ $F_"frict"=mumg=4xx5xx9.8=196$ $N$ $N$

To accelerate a mass the force given by Newton's Second Law is:

$F_{accel} = m a = 5 \times 6 = 30$ $F_"accel"=ma=5xx6=30$ $N$ $N$

Then:

$F = F_{frict} + F_{accel} = 196 + 30 = 226$ $F=F_"frict"+F_"accel"=196+30=226$ $N$ $N$

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An object with a mass of $5$ $5$ $k g$ $kg$ is on a surface with a kinetic friction coefficient of $4$ $4$ . How much force is necessary to accelerate the object horizontally at $6$ $6$ $m s^{- 2}$ $ms^-2$ ?

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An object with a mass of $5$ $5$ $k g$ $kg$ is on a surface with a kinetic friction coefficient of $4$ $4$ . How much force is necessary to accelerate the object horizontally at $6$ $6$ $m s^{- 2}$ $ms^-2$ ?