Question #33a3c

2 Answers
Aug 26, 2016

Answer:

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#

Aug 27, 2016

Answer:

#=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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cheg modified

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#