An object with a mass of $7 k g$ $7 kg$ is on a surface with a kinetic friction coefficient of $8$ $8$ . How much force is necessary to accelerate the object horizontally at $1 \frac{m}{s^{2}}$ $1 m/s^2$ ?

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Answer 1 · 2018-05-10T19:16:42

$555.8 N$ $"555.8 N"$

On a frictionless surface

$F = ma$ $"F = ma"$

Here, surface is rough. So, an additional force is required to have same acceleration $a$ $"a"$ . This additional force is called frictional Force.

$F = ma + μmg$ $"F = ma + μmg"$

$F = ({7 kg \times 1 m/s}^{2}) + (8 \times {7 kg \times 9.8 m/s}^{2}) = 555.8 N$ $"F" = ("7 kg × 1 m/s"^2) + (8 × "7 kg × 9.8 m/s"^2) = "555.8 N"$

An object with a mass of 7 kg7kg7 kg is on a surface with a kinetic friction coefficient of 8 8 8 . How much force is necessary to accelerate the object horizontally at 1 m/s^21ms2 1 m/s^2?