If an object is moving at #4 ms^-1# over a surface with a kinetic friction coefficient of #mu_k=16 /g#, how far will the object continue to move?

Question

If an object is moving at #4 ms^-1# over a surface with a kinetic friction coefficient of #mu_k=16  /g#, how far will the object continue to move?

Olivia Askew · Accepted Answer

The object will decelerate (accelerate in a direction opposite to its velocity) and stop after #1/2 m# due to the frictional force acting on it.

If there were no friction, the object would keep moving together. Since there is friction, there is a net unbalanced force acting that will cause the object to slow and stop. Summarising what we know and what we need to know: #mu_k=16/g# - coefficient of kinetic friction #u = 4ms^-1# - initial velocity #v = 0ms^-1# - final velocity #d = ? m# - distance to stop We will need to calculate the acceleration of the box, #a (ms^2)#, to calculate the distance. We use Newton's Second Law : #a=F/m# Now the force will just be the frictional force, which is the frictional coefficient times the weight force of the object, #mg#: #F_f = F_(weight)*mu_k = mg*16/g = 16 m# (g cancels) Substituting into Newton's Second Law: #a = F_f/m = (16m)/m = 16 ms^-2# Now we have #v, u, a# and #d#, so we will use: #v^2 = u^2 + 2ad# Rearranging to make #d# the subject: #d = (v^2 - u^2)/2a = (0^2 - 4^2)/(2*16) = 1/2 m# So the object will stop after #1/2m# or #0.5m#. (There is an issue with this question: a coefficient of friction should be 'dimensionless' (have no units), but #g# has the units of #ms^-2#. It's on the bottom line so the units over all would be #m^-1s^2#. I think what the person who set the question was trying to do was have #g# cancel out in the calculation, as you saw above, but it's poor physics. A coefficient of friction is just a number.)

Lily Adams · Answer

I found #0.5m#

From Newton's Second Law, #SigmavecF=mveca#, you can see that the object will receive an acceleration opposite to the direction of movement (deceleration) caused by friction #f_k=mu_k*N# (where in this case the normal reaction #N# is equal to the weight and given as: #N=mg#) and given as:

#-f_k=ma#

so that: #a=-f_k/m=-(mu_k*N)/m=-(mu_k*mg)/m=-mu_k*g=# #=-16/g*g=-16m/s^2#

with this information we can use the kinematic relationship:

#v_f^2=v_i^2+2ad# where the final velocity will be zero: #0^2=4^2-2*16d# and #d=0.5m#

Adrian Ayala · Answer

undefined To calculate the distance the object will continue to move, you can use the formula:

$$ d = \frac{{v^2}}{{2 \cdot \mu_k \cdot g}} $$

Where:
- $ d $ is the distance traveled,
- $ v $ is the initial velocity of the object (4 m/s in this case),
- $ \mu_k $ is the coefficient of kinetic friction (0.16 in this case),
- $ g $ is the acceleration due to gravity (approximately $ 9.81 \, \text{m/s}^2 $).

Substitute the values into the formula and solve for $ d $.

$$ d = \frac{{(4 \, \text{m/s})^2}}{{2 \cdot (0.16) \cdot (9.81 \, \text{m/s}^2)}} $$