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 $12 \frac{m}{s^{2}}$ $12 m/s^2$ ?

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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 $12 \frac{m}{s^{2}}$ $12 m/s^2$ ?

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Answer 1 · 2016-04-16T09:41:01

Net force = $F_{n} = m \times a = 7 k g \times 12 \frac{m}{s^{2}} = 84 N$ $F_n=mxxa=7kgxx12m/s^2= 84 N$
Frictional force = $F_{F} = μ m g = 3 \times 7 k g \times 9.8 \frac{m}{s^{2}} = 205.8 N$ $F_F=mumg=3xx7kgxx9.8m/s^2=205.8N$
So the applied force to have necessary acceleration $F_{a p p} = F_{n} + F_{F} = 84 + 205.8 = 289.8 N$ $F_(app)=F_n+F_F=84+205.8=289.8N$

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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 $12 \frac{m}{s^{2}}$ $12 m/s^2$ ?

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An object with a mass of 7kg7 kg is on a surface with a kinetic friction coefficient of 8 8 . How much force is necessary to accelerate the object horizontally at 12ms2 12 m/s^2?

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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 $12 \frac{m}{s^{2}}$ $12 m/s^2$ ?