a 68.5kg skater moving initially at 2.40 m/s on roughhorizontal ice comes to res

Question

a 68.5kg skater moving initially at 2.40 m/s on roughhorizontal ice comes to rest uniformly in 3.52s due to frictionfrom the ice. What force does friction exert on the skater? Answer is in N. M = 68.5 kg, V0=2.40m/s g = 9.80m/s2 and t=3.52s....I am obviously not settingequation up correctly. a 68.5kg skater moving initially at 2.40 m/s on roughhorizontal ice comes to rest uniformly in 3.52s due to frictionfrom the ice. What force does friction exert on the skater? Answer is in N. M = 68.5 kg, V0=2.40m/s g = 9.80m/s2 and t=3.52s....I am obviously not settingequation up correctly.

Explanation / Answer

Given that The mass of the skater (m) =68.5kg The initial velocity (u) =2.40m/s The final velocity becomes (v) =0m/s The time taken for the skater to come to rest (t) =3.52s Now we know that                     F = ma                          =m(v-u)/t Now substitute all the above values you get the required force does friction exert on the skater                   

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