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

`v=7.62\ m.s^(-1)`

Step-by-step 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=(F)/(m)`

Now force change in force on the body:

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

`F=18-0.53* (14-0)`

`F=10.58\ N`

Now the acceleration due to the force:

`a=(F)/(m)`

`a=(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* 2.0745* 14`

`v=7.62\ m.s^(-1)`