Question

The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. Find
a) The acceleration of the masses
1.8 m/s^2
2.4 m/s^2
3.2 m/s^2
4.1 m/s^2
b) The tension in the rope.
18 N
26 N
38 N
44 N

Answer 1

The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

T - 25 N = (8 kg) a

where T is the tension in the rope and a is the acceleration.

The hanging mass has a net force of

(6 kg) g - T = (6 kg) a

where g = 9.8 m/s².

Adding these equations together eliminates T, and we can solve for a :

(T - 25 N) + ((6 kg) g - T ) = (14 kg) a

33.8 N = (14 kg) a

a = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

T - 25 N = (8 kg) (2.4 m/s²)

T ≈ 25 N + 19.31 N ≈ 44 N