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

d = 0.391 m

Step-by-step 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

`(1)/(2)mv^2- Fd - mgh = 0`

`(1)/(2)mv^2- (\mu mg cos\theta )d - mgdsin\theta = 0`

`(1)/(2)mv^2 = (\mu mg cos\theta )d + mgdsin\theta`

`(1)/(2)mv^2 = d mg (\mu cos\theta+sin\theta)`

`d = (v^2)/(2g (\mu cos\theta+sin\theta))`

`d = (2.3^2)/(2* 9.8 (0.25* cos28^0+sin28^0))`

d = 0.391 m

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