Aforce table is set up so there is a pulley at 90° and one at 300°, with one free to be placed anywhere. the tension in the thread on the first pulley (90°) is 15 n. the tension in the thread on the second pulley (300°) is 10 n. what is the tension in the final thread and where should the pulley be placed?

The tension in the final thread is 20.41 N and the angle is 51.73°

Explanation:

Given that,

First pulley at  90°

Second pulley at 300°

Tension on first pulley = 15 N

Tension on second pulley = 10 N

We need to calculate the tension in the final thread

Using formula of resultant

We need to calculate the angle

Using formula of angle

Put the value into the formula

The pulley should be placed at 51.73°.

Hence, The tension in the final thread is 20.41 N and the angle is 51.73°

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s