If the coefficient of kinetic friction between


If the coefficient of kinetic friction between the 50-kg crate and the ground is \mu_k=0.3, determine the distance the crate travels and its velocity when t=3 s. The crate starts from rest, and P = 200 N.

If the coefficient of kinetic friction between

Image from: R. C. Hibbeler, K. B. Yap, and S. C. Fan, Mechanics for Engineers: Dynamics (SI Edition), 13th ed. Singapore: Pearson Education South Asia, 2013.

Solution:

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Let us first draw a free body diagram. Remember that friction is always opposite to the direction the object moves.

If the coefficient of kinetic friction between

N_f is the normal force, which is always perpendicular to the object.

Let us now write our equations of motion. Note that since the crate is only moving in the x-direction (horizontal), there is no acceleration in the y-direction (vertical), thus a_y=0.

+\uparrow \sum F_y=0;

N_f-490.5+200\sin30^0=0

N_f=390.5 N

 
For the x-axis, we have the following:

\rightarrow^+ \sum F_x=ma_x

200\cos30^0-0.3(N_f)=(50)(a)

200\cos30^0-0.3(390.5)=(50)(a)

a=1.12m/s^2

 

Now that we have the acceleration, we can figure out the velocity and the distance traveled. Note that the acceleration of the crate is constant thus, we can use our kinematics equations.

v=v_0+at

(Where v is final velocity, v_0 is initial velocity, a is acceleration and t is time)

 
Substitute the values we know:

v=v_0+at

v=0+1.12(3)

v=3.36 m/s

 

To find the distance, use the following formula:

s=s_0+v_0t+\dfrac{1}{2}at^2

(Where s is final displacement, s_0 is initial displacement, v_0 is initial velocity, t is time, and a is constant acceleration)

 

Substitute the values we know:

s=s_0+v_0t+\dfrac{1}{2}at^2

s=0+0(3)+\dfrac{1}{2}(1.12)(3)^2

s=5.04 m

 

Final Answers:

v=3.36 m/s

s=5.04 m

 

This question can be found in Engineering Mechanics: Dynamics (SI edition), 13th edition, chapter 13, question 13-3.

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