Answer:
10.86 N
Explanation:
Let the average frictional force acting on the toboggan be 'f' N.
Given:
Mass of toboggan (m) = 19.0 kg
Initial velocity (u) = 4.00 m/s
Final velocity (v) = 0 m/s
Time for which friction acts (Δt) = 7.00 s
Now, change in momentum is given as:
[tex]\Delta p =Final\ momentum-Initial\ momentum\\\\\Delta p=mv-mu\\\\\Delta p=19.0\ kg(0-4.00)\ m/s\\\\\Delta p=-76.00\ Ns[/tex]
Now, we know that, change in momentum is equal to the impulse acting on the body. So,
Impulse is, [tex]J=\Delta p=-76.00\ Ns[/tex]
Now, we know that, impulse is also given as the product of average force and the time interval for which it acts. So,
[tex]J=f\times \Delta t[/tex]
Rewriting the above equation in terms of 'f', we get:
[tex]f=\dfrac{J}{\Delta t}[/tex]
Plug in the given values and solve for 'f'. This gives,
[tex]f=\frac{-76.00\ Ns}{7.00\ s}\\\\f=-10.86\ N[/tex]
Therefore, the magnitude of frictional force is [tex]|f|=|-10.86\ N|=10.86\ N[/tex]
