Find acceleration of sled when child releases the rope with the tension being 16 N.

Question

A child pulls a 3.6 kg sled at 25 degrees to a slope that is 15 degrees to the horizontal. The sled moves at a constant velocity when the tension is 16 N. What is the acceleration of the sled if the rope is released?

Answer 1

There are 2 parts in the problem.

Part 1.

The child is pulling the sled up at angle α = 25° to a slope that is β = 15º to the horizontal. In this case 4 forces acts on the sled:

1. weight mg vertically down
2. normal reaction force N₁ perpendicular to the incline
3. tension T at angle α = 25° to a slope
4. frictional force f down and parallel to the plane (without presence of this force it will be impossible to pull the sled with constant speed).

Let's select the coordinate system in a way that x-axes directed upward and parallel to the surface of the incline. In scalar form the second Newton's law could be written as:

Tcosα - f - mgsinβ = 0 - we don't have acceleration, because the sled moves with constant velocity)

From this equation we can find friction f = Tcosα - mgsinβ .

After plugging in the values of variables to this equation we will get:

f = 16 N x cos 25° - 3.6 kg x 9.8 m/s² x sin 15º = 5.4 N

Part 2.

Now the sled is moving down after the child has released the rope. Now only 3 forces act on the sled:

1. weight mg vertically down
2. normal reaction force N₂ perpendicular to the surface of the incline
3. friction force f with the magnitude of 5.4 N parallel to the incline, but in this case upward.

For this case let's direct the x-axis parallel to the surface of the incline and down. In scalar form the Newton's second law looks like this:

mgsinβ - f = ma

From this equation we express acceleration

a = ( mgsinβ - f)/m

After substitution of all numbers, we will get:

a = (3.6 kg x 9.8 m/s² x sin 15° - 5.4 N)/3.6 kg = 1.0 m/s²

The answer: 1.0 m/s²

Alara, I hope that you understood the solution. Sorry, that cannot show you the pictures.