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 # 21 m/s^2#?

1 Answer
Jun 16, 2016

Answer:

#700N#

Explanation:

Newton's second law of motion states that the sum of the forces acting on an object is equal to its mass multiplied by its acceleration.

Mathematically speaking,

#color(blue)(|bar(ul(color(white)(a/a)F_"net"=macolor(white)(a/a)|)))#

where:
#F_"net"=#net force
#m=#mass #(kg)#
#a=#acceleration #(m/s^2)#

In your case, we will start off by letting the up and forward directions be positive.

Step 1
List out all the forces acting on the object.

#F_N+F_g+F_(app)+F_f=ma#

Since the object is not changing in the #y# direction, or vertically, the normal and gravity forces cancel each other out.

#color(red)cancelcolor(black)(F_N)+color(red)cancelcolor(black)(F_g)+F_(app)+F_f=ma#

So we are left with,

#F_(app)+F_f=ma#

Step 2
Since you are looking for the force necessary to accelerate the object, isolate for #F_(app)#.

#F_(app)=ma-F_f#

Step 3
In the equation, #F_f# can be simplified into

#F_(app)=ma-mu_kF_N#

Note: Since the object is not changing vertically, #F_N=F_g=mg#.

#F_(app)=ma-mu_kmg#

Factoring out #m#,

#F_(app)=m(a-mu_kg)#

Step 4
Plug in your known values.

#F_(app)=(7kg)[21m/s^2-(8)(-9.81m/s^2)]#

Solve.

#F_(app)=696.36N#

Rounding off the answer to one significant figure,

#F_(app)~~color(green)(|bar(ul(color(white)(a/a)color(black)(700N)color(white)(a/a)|)))#