Question

A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 5.10-kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box.If the box is initially at rest at x=0, what is its speed after it has traveled 14.0 m ?

Answer 1

Answer:

$v=7.62\ m.s^{-1}$

Explanation:

Given:

• initial position of the box, $x=0\ m$

• final position of the box, $x'=14\ m$

• mass of the box under the force, $m=5.1\ m$

• initial speed of the box, $u=0\ m.s^{-1}$

• function of force, $F(x)=18-0.53x\ [N]$

where:

$x=$ distance in the +ve x-direction

We know:

$F=m.a\\\Rightarrow a=\frac{F}{m}$

Now force change in force on the body:

$F(x)=18-0.53(x'-x)$

$F=18-0.53\times (14-0)$

$F=10.58\ N$

Now the acceleration due to the force:

$a=\frac{F}{m}$

$a=\frac{10.58}{5.1}$

$a=2.0745\ m.s^{-2}$

Now using equation of motion:

$v^2=u^2+2a.x'$

$v^2=0^2+2\times 2.0745\times 14$

$v=7.62\ m.s^{-1}$