Question

Two packing crates of masses 10.0 kg and 5.00 kg are connected by a light string that passes over a frictionless pulley. The 5.00 kg crate lies on a smooth incline ofangle 40.0º. Find the acceleration of the 5.00 kg crate and the tension in the sting.

Answer 1

Downward Force on 10 kg side = 10*g
Downward force on 5 kg side = 5*g*sin40

let acceleration of system be a

(10+5)*a = 10g - 5gsin40
=> a = 4.43

Now force on 10 kg block is 10g and tension

98 -T= 10*4.43
=> T = 53.7N