Question

A 18 kg sled starts up a 28 degree incline with a speed of 2.3 m/s. The coefficient of kinetic friction between the sled and the incline is 0.25. How far up the incline will the sled travel before coming to a stop?

Answer 1

Answer:

d = 0.391 m

Explanation:

given,

mass of sled = 18 kg

inclined at an angle of  = 28°

kinetic friction between the sled and inclined = 0.25

using energy equation

$\dfrac{1}{2}mv^2- Fd - mgh = 0$

$\dfrac{1}{2}mv^2- (\mu mg cos\theta )d - mgdsin\theta = 0$

$\dfrac{1}{2}mv^2 = (\mu mg cos\theta )d + mgdsin\theta$

$\dfrac{1}{2}mv^2 = d mg (\mu cos\theta+sin\theta)$

$d = \dfrac{v^2}{2g (\mu cos\theta+sin\theta)}$

$d = \dfrac{2.3^2}{2\times 9.8 (0.25\times cos28^0+sin28^0)}$

d = 0.391 m

hence, the distance traveled by the sled is equal to 0.391 m