Find the Difference in Tension: Steel Cable & 20kg Block

[madskier8806]

Homework Statement

A massive steel cable drags a 20.0 kg block across a horizontal, frictionless surface. A 90.0 N force applied to the cable causes the block to reach a speed of 3.60 m/s in 2.40 s.
What is the difference in tension between the two ends of the cable?

Homework Equations

F=ma
a=v/t

The Attempt at a Solution

a=3.6/2.4= 1.5 m/s2

90N=m*1.5
m=60
m(block)=20kg
m(cable)=40kg

F=40*1.5=60

90-60=30

Just looking at the numbers it doesn't look right

Thanks all for your help

 

[PhanthomJay]

Homework Statement

A massive steel cable drags a 20.0 kg block across a horizontal, frictionless surface. A 90.0 N force applied to the cable causes the block to reach a speed of 3.60 m/s in 2.40 s.
What is the difference in tension between the two ends of the cable?

Homework Equations

F=ma
a=v/t

The Attempt at a Solution

a=3.6/2.4= 1.5 m/s2yes

90N=m*1.5
m=60
m(block)=20kg
m(cable)=40kgyes

F=40*1.5=60 this is F_net

90-60=30no

Just looking at the numbers it doesn't look right

Thanks all for your help

looks like you isolated the rope and came up with the NET force on the rope. Try drawing a FBD of the block. Or identify all forces acting on the rope.

 

[m2003]

I would like to clarify a few points in the given scenario. First, it is important to note that tension is a force, not a mass. Therefore, the statement "difference in tension between the two ends of the cable" is a bit unclear. I will assume that it refers to the difference in the magnitude of tension at the two ends of the cable.

Now, let's break down the problem and approach it systematically. We have a 20 kg block that is being dragged by a massive steel cable, which is being pulled by a 90 N force. We are given the speed and time, and we are asked to find the difference in tension between the two ends of the cable.

First, let's consider the block. From the given information, we know that it is accelerating at a rate of 1.5 m/s^2 (calculated correctly in the attempt at a solution). Using Newton's second law, we can calculate the net force acting on the block:

F = ma = (20 kg)(1.5 m/s^2) = 30 N

This means that there must be a force of 30 N acting on the block in the direction of motion. Since the surface is frictionless, this force must be provided by the tension in the cable.

Next, let's consider the cable. Since it is being pulled by a 90 N force, the tension in the cable must also be 90 N. However, as the block moves, the tension in the cable changes. At the end where the force is being applied, the tension must be greater than 90 N in order to accelerate the block. At the other end, the tension must be less than 90 N since the block is moving away from it.

Therefore, the difference in tension between the two ends of the cable is the difference between the maximum and minimum tension values. Using the calculated force of 30 N on the block, we can set up the following equations:

T1 + 30 N = 90 N
T2 - 30 N = 0 N

where T1 is the tension at the end where the force is being applied and T2 is the tension at the other end. Solving these equations, we get:

T1 = 60 N
T2 = 30 N

The difference in tension is then:

T1 - T2 = 60 N - 30 N = 30 N