If a #2 kg# object moving at #5 m/s# slows down to a halt after moving #3 m#, what is the friction coefficient of the surface that the object was moving over?

Question

Julia Aguilar · Accepted Answer

#u_k~=0,425# #E_k=1/2*m*v^2 " The kinetic energy of object"# #W=F_f* Delta x " Work doing by friction force"# #F_f=u_k*N " N: normal force to contacting surfaces"# #"The kinetic energy turns work"# #1/2*cancel(m)*v^2=u_k.cancel(m)*g*Delta x# #1/2*5^2=u_k*9,81*3# #25=u_k*2*3*9,81# #u_k=25/(2.3.9,81)# #u_k~=0,425#

Sebastian Acosta · Answer

undefined To find the friction coefficient, we can use the equation:

$$ F_{\text{friction}} = \mu \times F_{\text{normal}} $$

where:
- $ F_{\text{friction}} $ is the force of friction
- $ \mu $ is the coefficient of friction
- $ F_{\text{normal}} $ is the normal force

The force of friction can also be expressed as:

$$ F_{\text{friction}} = m \times a $$

where:
- $ m $ is the mass of the object
- $ a $ is the acceleration

Given:
- $ m = 2 $ kg
- initial velocity, $ v_0 = 5 $ m/s
- final velocity, $ v = 0 $ m/s (object comes to a halt)
- distance traveled, $ d = 3 $ m

Using the equation for acceleration:

$$ a = \frac{{v^2 - v_0^2}}{{2d}} $$

Substituting the given values:

$$ a = \frac{{0^2 - 5^2}}{{2 \times 3}} = \frac{{-25}}{{6}} $$

The negative sign indicates deceleration.

Now, using $ F_{\text{friction}} = m \times a $, we find:

$$ F_{\text{friction}} = 2 \times \frac{{-25}}{{6}} = -\frac{{25}}{{3}} $$

Since the force of friction opposes the motion, it is negative.

Finally, using $ F_{\text{friction}} = \mu \times F_{\text{normal}} $, and since the object comes to a halt, the force of friction is equal to the force applied:

$$ \mu \times m \times g = \frac{{25}}{{3}} $$

$$ \mu = \frac{{\frac{{25}}{{3}}}}{{m \times g}} $$

$$ \mu = \frac{{\frac{{25}}{{3}}}}{{2 \times 9.8}} $$

$$ \mu \approx 0.426 $$

Therefore, the friction coefficient of the surface is approximately $ 0.426 $.