Question

A 250 kg cart starts from rest and rolls down an inclined plane from a height of 550 m. Determine its speed at a height of 125 m above the bottom of the incline. Please round to two decimal places.

Answer 1

24.85 m/s.

Step-by-step explanation:

PE = mgh = 250 kg x 9.81 m/s^2 x 550 m = 1,358,725 J

PE' = mgh' = 250 kg x 9.81 m/s^2 x 125 m = 308,062.5 J

PE = KE

1,358,725 J = 0.5mv^2

Solving for v, we get:

v = sqrt(2PE/m) = sqrt(2 x 1,358,725 J / 250 kg) = 59.15 m/s (rounded to two decimal places)

PE' + KE' = PE + KE

Since the cart starts from rest at the top of the incline, KE = 0. Therefore:

PE' = KE'

mgh' = 0.5mv'^2

Solving for v', we get:

v' = sqrt(2gh') = sqrt(2 x 9.81 m/s^2 x 125 m) = 24.85 m/s (rounded to two decimal places)

Therefore, the speed of the cart at a height of 125 m above the bottom of the incline is 24.85 m/s.