Can anyone help me with parts A and B?

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1 Answer
Jun 7, 2018

A. #t ~= 2.3 s#
B. #s ~= 5.6 m#

Explanation:

A. The first question is "does the box start to move wrt (with respect to) the truck bed?".

The force of maximum static friction, #F_"sf"#, is

#F_"sf" = mu_s*N = mu_s*m*g = 0.18*11.1 kg*9.8 m/s^2 = 19.58 N#

If the box is to remain stationary wrt the truck, the friction needs to be able to apply force of

#F = m*a = 11.1 kg*2.17 m/s^2 = 24.1 N# to the box. The maximum #F_"sf" = 19.58 N#, so the box will start sliding.

The second question is "what will the acceleration of the box be?"

#F_"kf" = mu_k*N = mu_k*m*g = 0.15*11.1 kg*9.8 m/s^2 = 16.32 N#

That force will accelerate the box forward wrt the road according to

#a_r = F/m = 16.32 N/11.1 kg = 1.47 m/s^2#

That is less than the truck's acceleration of 2.17 m/s^2. The difference between the 2 values of acceleration is

#a_Delta = 2.17 m/s^2 - 1.47 m/s^2 = 0.7 m/s^2#

Therefore, to the box it will seen that the edge of the back of the truck's deck is accelerating toward it at #0.7 m/s^2#. The edge of the truck bed was originally 1.81 m away. The final question is "how much time elapses before that has decreased to zero?"

#s = u*t + 1/2*a_Delta*t^2#

#1.81 m = 0 + 1/2*0.7 m/s^2*t^2#

#t^2 = (2*1.81 cancel(m))/(0.7 cancel(m)/s^2) = 5.17 s^2#

#t = 2.27 s ~= 2.3 s#

B. The distance the truck travels in 2.3 s is

#s = u*t + 1/2*a_"truck"*t^2#

#s = 0 + 1/2*2.17 m/s^2*(2.3 s)^2 = (2.17 m/cancel(s^2)*2.3^2 cancel(s^2))/2#

#s = (2.17 m*2.3^2 )/2 = 5.59 ~= 5.6 m#

I hope this helps,
Steve

P.S. Please ask me for any explanation you might need.