Question

a trolley of mass 5 kg is connected to block of mass 2 kg with the help of the massless inextensible string passes over a light frictionless pulley find the tension in the string
assume no friction
g=10m/s-²
sin@=4/5 and
cos@=3/5​

Answer 1

T=5a....(1)
20sin53'–T = 2a... (2)
from 1&2..
20x 4/5=7a
a=16/7
& T=5a =› T=5x16/7

Answer 2

Answer:

The tension in the string is 15.68 N.

Explanation:

Given that,

Mass of trolley = 5 kg

Mass of block = 2 kg

We need to calculate the tension in the string

According to FBD,

The tension is

$T= mg\sin\theta$

Where, m = mass of block

g = acceleration due to gravity

Put the value into the formula

$T=2\times9.8\times\dfrac{4}{5}$

$T=15.68\ N$

Hence, The tension in the string is 15.68 N.