An object with a mass of #5 kg# is on a surface with a kinetic friction coefficient of # 4 #. How much force is necessary to accelerate the object horizontally at # 17 m/s^2#?

Question

An object with a mass of #5  kg# is on a surface with a kinetic friction coefficient of # 4  #.  How much force is necessary to accelerate the object horizontally at #         17  m/s^2#?

Lily Adams · Accepted Answer

The force is #=281N#

The frictional force is

#F_r=mu_k*N#

The normal force is #N=mg#

So,

#F_r=mu_k*mg#

Resolving in the horizontal direction #rarr^+#

We apply Newton's second Law

#F-F_r=ma#

So,

#F=F_r+ma#

#F=mu_kmg+ma#

#=m(mu_kg+a)#

#=5(4*9.8+17)#

#=5*56.2=281N#

Benjamin Asiedu · Answer

undefined To calculate the force required to accelerate the object horizontally at 17 m/s^2, you can use the equation:

$$F_{\text{net}} = m \times a$$

Where:
- $F_{\text{net}}$ is the net force,
- $m$ is the mass of the object, and
- $a$ is the acceleration.

Given that the mass $m = 5$ kg and the acceleration $a = 17$ m/s^2, the net force ($F_{\text{net}}$) can be calculated. However, since there is kinetic friction, we need to account for it. The force of kinetic friction ($F_{\text{friction}}$) can be calculated using the equation:

$$F_{\text{friction}} = \mu_k \times N$$

Where:
- $\mu_k$ is the coefficient of kinetic friction, and
- $N$ is the normal force.

The normal force ($N$) can be calculated using the equation:

$$N = mg$$

Where:
- $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$).

After finding $F_{\text{friction}}$, subtract it from $F_{\text{net}}$ to get the total force required to accelerate the object.