To solve these tension word problems, you will mostly need Newton's second law of motion.

**Problem # 1**

a 100 kg bucket is being lifted by a rope. The bucket started at rest and after being lifted 4 m, it is moving at 4 m/s. If the acceleration is constant, what is the tension on the rope?

**Solution**

The diagram below is useful in order to visualize the problem.

**Observations**:

The bucket started from rest, so V

Then the bucket moved with a speed of 4 m/s, so V = 4 m/s

F

The positive direction is chosen to be up

F

F

F

F

The only missing quantity is a. To get a, we can use one of the constant acceleration equations

Since we know V4

16 = 0 + 8a

16 = 8a

a = 2 m/s

F

F

**Problem # 2:**

A rope is attached to a vehicle weighing 500 kN. A very strong man puts the other end of the rope inside his mouth at an angle of 30 degrees from the horizontal. If the strong man is able to move the vehicle with a constant force 1 m away with a speed of 0.25 m/s, what is the tension on the rope. (Use g = 10 m/s^2)

**Solution**

This situation is shown below with graph.

Newton's second law gives F = ma

Since the vehicle is only moving along the x axis, the only force on the vehicle is along the x axis.

We get FThe force F

cos(30°) =

T_{x}
T

T

T cos(30°) = ma

We need to find the acceleration and the mass of the vehicle.

m =

W
g

m =

500000
g

m =

500000
10

= 50000 kg
Since the man is moving the vehicle with a constant force, the acceleration is constant. Therefore, we can use the following equation again to get the acceleration.

V0.25

0.0625 = 0 + 2a

0.0625 = 2a

a = 0.03125 m/s

We can now find the tension on the rope

T cos(30°) = ma

T cos(30°) = 50000 × 0.03125

T cos(30°) = 1562.5

T × 0.866 = 1562.5

T = 1562.5 / 0.866

T = 1804.27 Newtons

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