Answer :
Answer:
[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]
Explanation:
For this case we can use the second law of Newton given by:
[tex] \sum F = ma[/tex]
The friction force on this case is defined as :
[tex] F_f = \mu_k N = \mu_k mg [/tex]
Where N represent the normal force, [tex]\mu_k [/tex] the kinetic friction coeffient and a the acceleration.
For this case we can assume that the only force is the friction force and we have:
[tex] F_f = ma[/tex]
Replacing the friction force we got:
[tex] \mu_k mg = ma[/tex]
We can cancel the mass and we have:
[tex] a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}[/tex]
And now we can use the following kinematic formula in order to find the distance travelled:
[tex] v^2_f = v^2_i - 2ad[/tex]
Assuming the final velocity is 0 we can find the distance like this:
[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]