A 54 N tow force accelerates a cord-joined pair over a frictionless floor: a 7.0 kg front box and a 2.0 kg rear box.
Two unknowns hide in the scene: the pair's shared acceleration, and the cord's tension. One equation used twice, on two different choices of object, gets both.
The trick is choosing what to call the object.
First, treat the whole pair as one object. The cord then acts inside the chosen object, between its own parts, so its tension drops out of the net force. One external force and one total mass remain.
Step 1: apply a = F_net/m to the whole system. The tow force is the only external horizontal force, and the mass is both boxes together.
For the pair: a = 54/(7.0 + 2.0) = 54/9.0. So a = 6.0 m/s², shared by both boxes.
Step 2: apply a = F_net/m to one box alone to find the tension. Choose the rear box, because the cord's tension is the only horizontal force on it.
For the rear box, F_net = ma means T = 2.0 × 6.0. So T = 12 N.
The order matters: the system step needs no tension, and the single-box step needs the acceleration the system step produced. Whole system first, then one object.
A sense check: the tension is less than the tow force. The cord only has to accelerate the rear box, never the whole pair.
A 54 N tow force accelerates a cord-joined pair over a frictionless floor: a 7.0 kg front box and a 2.0 kg rear box. Calculate the shared acceleration and the cord's tension.
Write down the values in the question:
tow force = 54 N
front box: 7.0 kg
rear box: 2.0 kg
a = ?
T = ?
Apply a = F_net/m to the whole system:
a = 54/(7.0 + 2.0)
a = 54/9.0
a = 6.0 m/s²
Apply a = F_net/m to the rear box alone, where the tension is the only horizontal force:
T = ma
T = 2.0 × 6.0
T = 12 N
The shared acceleration is 6.0 m/s², and the cord's tension is 12 N.
A 33 N tow force accelerates a cord-joined pair over a frictionless floor: a 4.0 kg front box and a 7.0 kg rear box. Calculate the shared acceleration and the cord's tension.
Write down the values in the question:
tow force = 33 N
front box: 4.0 kg
rear box: 7.0 kg
a = ?
T = ?
Apply a = F_net/m to the whole system:
a = 33/(4.0 + 7.0)
a = 33/11
a = 3.0 m/s²
Apply a = F_net/m to the rear box alone, where the tension is the only horizontal force:
T = ma
T = 7.0 × 3.0
T = 21 N
The shared acceleration is 3.0 m/s², and the cord's tension is 21 N.
A 60 N tow force accelerates a cord-joined pair over a frictionless floor: an 8.0 kg front crate and a 4.0 kg rear crate. Enter the pair's shared acceleration in m/s².
The same pair: the 8.0 kg and 4.0 kg crates accelerate together at 5.0 m/s². The cord's tension is the only horizontal force on the rear crate. Enter the cord's tension in N.
A 45 N tow force accelerates a cord-joined pair over a frictionless floor: a 6.0 kg front box and a 3.0 kg rear box. Enter the cord's tension in N.
A 63 N tow force accelerates a cord-joined pair over a frictionless floor: an 8.0 kg front sled and a 6.0 kg rear sled. Enter the pair's shared acceleration in m/s².
A 56 N tow force accelerates a cord-joined pair over a frictionless floor: a 4.0 kg front bin and a 3.0 kg rear bin. Enter the cord's tension in N.
At a boatyard, a winch line applies a 90 N force to the front dolly of a cord-joined pair rolling on a frictionless track. The front dolly is 12 kg and the rear dolly is 6.0 kg. Enter the cord's tension in N.