Question

3. A trolley accelerates from rest at point A. down a straight track to point B. The trolley then continues along a horizontal track to point C, as shown. The effects of air resistance and friction are negligible. Given that height of the trolley = 0.50 m and A-B in straight line= 2.0 m

(a) Show that the trolley reaches point B with a speed of about 3 m/s​

Answer 1

By using conservation of energy and the negligible effects of air resistance and friction, the trolley's potential energy at point A is converted into kinetic energy at point B, leading to a final speed of about 3 m/s.

Step-by-step explanation:

Calculating the Speed of a Trolley at Point B

To show that the trolley reaches point B with a speed of about 3 m/s, we can use the principles of energy conservation. Since air resistance and friction are negligible, the mechanical energy of the trolley is conserved. We'll assume the potential energy of the trolley at point A is converted into kinetic energy at point B.

The potential energy at point A (PEA) is equal to the product of mass m, gravitational acceleration g (approximately 9.8 m/s²), and height h. The kinetic energy (KEB) at point B is ½mv², where v is the velocity at point B.

Using the formula: PEA = KEB, we get:

mgh = ½mv²

Since the mass m cancels out, we can solve for v:

gh = ½v²

Substituting g = 9.8 m/s² and h = 0.50 m gives:

(9.8 m/s²)(0.50 m) = ½v²

4.9 m²/s² = ½v²

v² = 9.8 m²/s²

v = √(9.8 m²/s²)

v ≈ 3.13 m/s

Thus, the trolley reaches point B with a speed of approximately 3.13 m/s, which is about 3 m/s.

Answer 2

Step-by-step explanation:

Potential energy at A = kinetic energy at B

PE = KE

mgh = ½ mv²

gh = ½ v²

v = √(2gh)

Plug in values:

v = √(2 × 9.8 m/s² × 0.50 m)

v = 3.1 m/s