Answer to Question #189859 in Physics for nav

Question

On an icy road, a 10-kg cart is given a push that imparts to it an initial speed of v₀ = 2.0 m/s. The coefficient of kinetic friction between cart and ice is k = 0.10. Find the distance the cart moves before coming to rest.

Answer 1

The initial kinetic energy of the cart is:

$$K_i = \dfrac{mv_0^2}{2}$$

where $m = 10kg$ is the mass of the cart, and $v_0 = 2m/s$ is its initial speed. The final kinetic energy is $K_f = 0$, since the cart stops.

According to the work-energy theorem, the total work done by friction is:

$$W = -\Delta K = -(K_f - K_i) = \dfrac{mv_0^2}{2}$$

On the other hand, by definition, the work is:

$$W = F_{fr}\cdot d$$

where $d$ is the distance the cart moves before coming to rest, and $F_{fr}$ is the friction force. Int turn the friction force is defined as follows:

$$F_{fr} = k N = k mg$$

where $k = 0.1$ is the coefficient of kinetic friction between cart and ice, and $N = mg$ is the normal force (equal to the weight of the car, since it does not move verticaly). Finally, obtain:

$$k mg\cdot d = \dfrac{mv_0^2}{2}\\ d = \dfrac{mv_0^2}{2kmg} = \dfrac{v_0^2}{2kg} \\ d = \dfrac{2^2}{2\cdot 0.1\cdot 9.8} \approx 2.0m$$

Answer. 2 m.