Question

Question #33a3c

Answer 1

Well.

Explanation:

There is only a downward force and no upward force so we will focus there.

`sum F_x` = `m*g*sintheta + 26.0N - f_k`
`sum F_x` = `9kg*9.8 (m)/(s^2)*0.54+26.0N-[0.3*9kg*9.8 (m)/(s^2)*0.83]`
`sum F_x` = `47.6+26N-21.961N`
`sum F_x` = `51.64N`

Now, you are asked to find the velocity after t = 2 s and you know that the intial v is 0 since the box started from rest. You are going to have to use 1 of you kinematic equations

`v_f=v_o + a*t`

`v_o = 0`
`t = 2 s`
`v_f = ?`
`a = ?`

How do you find acceleration? Well you have found the net downward Force so using Newton's 2nd law of motion

`F=m*a`

`51.64N` = `9 kg`*`a`

`(51.64N)/(9kg)` = `a`

`a` = `5.73 (m)/(s^2)`

`v_f=v_o + a*t`
`v_f=0` + `5.73 (m)/(s^2)`*`2s`

`v_f= 11.46 m/s`

Answer 2

`=11.532ms^-1`

Explanation:

Question
A 9.00-kg box sits on a ramp that is inclined at 33.00 above the horizontal. The coefficient of friction between the box and the surface of the ramp is 0.300. A constant horizontal force F = 26.0 N is applied to the box (as in Figure given below) , and the box moves down the ramp.If the box is initially at rest, what is its speed 2.00 s after the force is applied?
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Answer

It is clear from the figure that the vertical component of applied force`Fsintheta` will diminish normal reaction but its horizontal component `Fcostheta`increases the downward force.parallel to the plane of the ramp.
So
`"Normal Reaction " N=mgcostheta-Fsintheta`

`"Frictional force " f=muN`

`"Net Downward force parallel to ramp"`

`=mgsintheta+Fcostheta-muN`

`=mgsintheta+Fcostheta-mu(mgcostheta-Fsintheta)`

`"Downward acceleration"`

`a =1/m(mgsintheta+Fcostheta-mu(mgcostheta-Fsintheta))`

`a=gsintheta+F/mcostheta-mu(gcostheta-F/msintheta)`

Inserting
`m=9kg, g=9.8ms^-2,mu=0.3,F=26N,theta=33^@`

`=>a=9.8sin33+26/9cos33-0.3(9.8cos33-26/9sin33)`

`=>a=5.337+2.423-0.3(8.219-1.573)ms^-2`

`=>a=5.337+2.423-1.994ms^-2~~5.766ms^-2 color(red)(" rounded up to 3 decimal place")`

Now calculation of velocity 2s after the application of force F

`v_i->"Initial velocity"=0`

`a->"Acceleration"=5.766ms^-2`

`t->"Time"=2s`

`v_f->"Final velocity"=?`

`v_f=v_i+axxt`

`=>v_f=0+5.766xx2=11.532ms^-1`