A 5 kg block slides along a course horizontal surface and slows to 6.5 m/s after traveling 40 m. If the kinetic coefficient of friction between the block and surface is 0.23, the intial speed (m/s) of the block was closest to:

1 Answer

James Dowd

Updated on January 2nd, 2021

In our problem, we first want to understand the forces acting upon the object. We know that the force of gravity will exist on the object. The force of gravity produces an equal and opposite force called the normal force. Finally, the force of friction will occur in the opposite direction that the block moves. 

Let's show this visually:

The only forces that will affect the block's acceleration would be the forces acting on it in the same axis of movement. Therefore, since the block is moving horizontally, the only force that would affect the acceleration is the force of friction. The force of kinetic friction (since our object is moving), can be found on pg. 122 of the FE Reference Handbook and is as follows:

Ffriction=μkN N =Normal force μk=Coefficient of kinetic friction

Since our normal force is equal in magnitude to the force of gravity, we can use the block's mass and the acceleration of gravity (g=9.8) to find the normal force:

Fg=mg Fg=(5kg)(9.8ms2) Fg=49N N = 49N

Plugging our normal force back into the equation of force of friction with the kinetic coefficient given in the problem:

Ffriction=μkN Ffriction=(.23)(49) Ffriction=11.27N

Therefore, the force of friction is acting 11.27 Newtons in the opposite direction of movement of the block. This will cause the block to deaccelerate, as described in the problem. Let's now find the acceleration of the block according to Newton's laws of motion:

F=ma 11.27N = (5kg)(a) a = 2.254m/s2

Since the acceleration is acting in the opposite direction of motion of the block, we can think of it as a negative acceleration, or -2.254m/s^2. Finally, we use the following equation to find the initial speed of the block based on the information given in the problem and what we have solved so far:

Vi=2(d  Vi=6.5-(2*(-2.254)(40) Vi=6.5-(2*(-2.254)(40) Vi=186.82 Vi=13.67ms

Therefore, our initial velocity is approximately 13.67m/s.

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