Question

At one particular moment, a 19.0 kg toboggan is moving over a horizontal surface of snow at 4.00 m/s. After 7.00 s have elapsed, the toboggan stops. Use a momentum approach to find the magnitude of the average friction force (in N) acting on the toboggan while it was moving.

Answer 1

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:

$\Delta p =Final\ momentum-Initial\ momentum\\\\\Delta p=mv-mu\\\\\Delta p=19.0\ kg(0-4.00)\ m/s\\\\\Delta p=-76.00\ Ns$

Now, we know that, change in momentum is equal to the impulse acting on the body. So,

Impulse is, $J=\Delta p=-76.00\ Ns$

Now, we know that, impulse is also given as the product of average force and the time interval for which it acts. So,

$J=f\times \Delta t$

Rewriting the above equation in terms of 'f', we get:

$f=\dfrac{J}{\Delta t}$

Plug in the given values and solve for 'f'. This gives,

$f=\frac{-76.00\ Ns}{7.00\ s}\\\\f=-10.86\ N$

Therefore, the magnitude of frictional force is $|f|=|-10.86\ N|=10.86\ N$