Answer to Question #185823 in Mechanics | Relativity for Ojo Claudius

Question

A body of mass 2 kg is moving on a smooth horizontal surface under the action of a horizontal force F=55+t² (N). Calculate the velocity of the body at t=5s, assuming the body was at rest at t=0. (Answer is 158.3m/s)

Answer 1

To be given in question

Mass =2kg

Force(F)=$55+t^2(N)$

t=5sec

To be asked in question

Velocity of body(V)=?

We know that

$F=ma=m\frac{dv}{dt}\rightarrow(1)$

Put value

$55+t^2=2\frac{dv}{dt}$

$\frac{dv}{dt}=\frac{1}{2}[55+t^2]\rightarrow(2)$

eqution (2)take integration with respect to time

$v=\frac{1}{2}[55t+\frac{t^3}{3}]\rightarrow(3)$

Put(t=5sec)in eqution (3)

$v=\frac{1}{2}[55\times5+\frac{125}{3}]$

$v=\frac{1}{2}[275+\frac{125}{3}]$

$v=\frac{950}{6}=158.33meter/sec$

$v=158.33meter/sec$