An object with a mass of 7 kg is on a surface with a kinetic friction coefficient of  8 . How much force is necessary to accelerate the object horizontally at  32 m/s^2?

Jan 14, 2018

$784 N$

Explanation: As,the object is going on a horizontal surface,frictional force acting on it will be $f = u \cdot N$ or $u m g$ i.e (8*7*10) N or, 560 N

So,le't's assume we will be requiring a force of F to cause an acceleration of 32 SI units on it.

so,we can write, $F - f = m \cdot a$(where, m is the mass of the object and a is its acceleration)

so, $F = \left(560 + 7 \cdot 32\right)$ N or, $784 N$

Jan 14, 2018

The force is $= 772.8 N$

Explanation:

The mass of the object is $m = 7 k g$

The acceleration is $a = 32 m {s}^{-} 2$

The coefficient of kinetic friction is

${\mu}_{k} = {F}_{r} / N = 8$

The normal force is $N = 7 g N$

The frictional force is ${F}_{r} = {\mu}_{k} \times N = 8 \cdot 7 g = 56 g N$

The force necessary to accelerate the object is $= F N$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

According to Newton's Second Law

$F - {F}_{r} = m a$

$F = m a + {F}_{r} = \left(\left(7 \times 32\right) + \left(56 g\right)\right) N = 772.8 N$