Discussion Forum

A solid horizontal surface is covered with  a thin layer of oil. A rectangular block of mass m= 0.4 kg  is at  rest on this surface . An impulse of 1.0 N  s is  applied to the block at time t=0, so  that it starts moving along the x- axis with a velocity $v_{t}=v_{0}e^{\frac{-t}{\tau}}$  where  v₀  is a constant  and   $\tau=4$ s. The displacement of the block, in metres at $t=\tau$ is...............  ( Take , e^-1  =0.37)

Answer:

Explanation:

Linear impluse, J= mv₀ 

   $v_{0}= \frac{J}{m}= 2.5m/s$

   $v_{t}=v_{0}e^{\frac{-t}{\tau}}$

  $\frac{\text{d}x}{\text{d}t}=v_{0}e^{-t/\tau}$

    $\int_{0}^{x}dx =v_{0}\int_{0}^{\tau}  e^{-t/\tau}dt$

  $x= v_{0}[\frac{e^{-t/\tau}}{\frac{-1}{\tau}}]_0^\tau$

       x= 2.5(-4)(e^-1 - e⁰ )

          = 2.5(-4)(0.37-1)

            x= 6.30 m