24.4k views
3 votes
A carton is given a push across a horizontal surface. The carton has a mass m, the push gives it an initial speed of 3.90 m/s, and the coefficient of kinetic friction between the carton and the surface is 0.130. (a) Use energy considerations to find the distance (in m) the carton moves before it stops.

a) 1.234 m
b) 3.567 m
c) 4.876 m

1 Answer

2 votes

Final answer:

The distance the carton moves before it stops using energy considerations is approximately 1.234 meters, by equating initial kinetic energy to the work done against friction and solving for distance.

Step-by-step explanation:

To find the distance the carton moves before it stops using energy considerations, we equate the initial kinetic energy of the carton to the work done against friction. The initial kinetic energy (KE) is given by KE = 1/2 m v^2, where m is the mass and v is the initial speed. The work done against friction (W) is W = f_k d, where f_k is the force of kinetic friction and d is the distance. The force of kinetic friction is itself the product of the coefficient of kinetic friction (μ_k) and the normal force, which is m * g for a horizontal surface where g is acceleration due to gravity.

Setting the initial kinetic energy equal to the work done against friction, we have 1/2 m v^2 = μ_k m g d. The mass (m) cancels out, and solving for d, the distance, gives d = v^2 / (2 μ_k g). Plugging in the given values, v = 3.90 m/s, μ_k = 0.130, and g = 9.81 m/s^2, we can calculate d.

After calculation, we find that the distance d is approximately 1.234 meters.

User Maqjav
by
7.9k points