Question

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 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 x=0, ​what is the speed after it has travelled 17.0m?

Answer 1

`v\approx 8.570\,(m)/(s)`

Step-by-step explanation:

The equation of equlibrium for the box is:

`\Sigma F_(x) = 18\,N-(0.530\,(N)/(m) )\cdot x = (7.90\,kg)\cdot a`

The formula for the acceleration, given in
`(m)/(s^(2))`, is:

`a = (18\,N-(0.530\,(N)/(m) )\cdot x)/(7.90\,kg)`

Velocity can be derived from the following definition of acceleration:

`a = v\cdot (dv)/(dx)`

`v\, dv = a\, dx`

`(1)/(2)\cdot v^(2) = \int\limits^(17\,m)_(0\,m) {(18\,N-(0.530\,(N)/(m) )\cdot x)/(7.90\,kg) } \, dx`

`(1)/(2)\cdot v^(2) =(18\,N)/(7.90\,kg) \int\limits^(17\,m)_(0\,m)\, dx - (0.530\,(N)/(m) )/(7.90\,kg) \int\limits^(17\,m)_(0\,m) {x} \, dx`

`(1)/(2)\cdot v^(2) = (2.278\,(m)/(s^(2)))\cdot x |_(0\,m)^(27\,m)-(0.034\,(1)/(s^(2)))\cdot x^(2)|_(0\,m)^(27\,m)`

`v =\sqrt_(0\,m)^(27\,m)-(0.034\,(1)/(s^(2)))\cdot x^(2)`

The speed after the box has travelled 17 meters is:

`v\approx 8.570\,(m)/(s)`