Answer to Question #3611 in Mechanics | Relativity for Vikash

Question

A force is applied on a body ,mass 50 kg, for 8 sec. After that the body cover 80m distance in 5 sec in coming to rest.Find the force applied.

Answer 1

Index “1” means first part of path and index “2” the second part
V is the velocity at time 8sec
F_f is a friction force

F₁ = F-F_f = ma₁
a₁ = (F-F_f)/m
V = a₁ t₂
F₂ = F_f = ma₂
a₂ = F₂/m
V = a₂ t₂ = F₂/m t₂ (*)
a₁ t₁ = a₂ t₂
(F-F_f)/m t₁ = F_f/m t₂
F = F_f (1+t₂/t₁ ) (**)
A₂ = F_f S₂
E₁ = (mV²)/2
A₂ = E₁
Let put here the V from (*)
F_f S₂ = m 1/2 F_f²/m² t₂²
so,
F_f = (2mS₂)/(t₂² )
(**):
F = (2mS₂)/(t₂² ) (1+t₁/t₂ ) = (2*50*80)/5² (1+8/5) = 832 N