Question

block with of mass m is at rest on horizontal frictionless surface at time t=0. A force given by F=Bt+C is applied horizontally to the center of gravity of the block. If m=4, B=C and C=0.5, what is the magnitude of block velocity at t=2 s?

Answer 1

`v_(2) =(1)/(2)`

Step-by-step explanation:

From the second law of Newton movement laws, we have:

`F=m*a`, and we know that a is the acceleration, which definition is:

`a=(dv)/(dt)`, so:

`F=m*(dv)/(dt)\\(dv)/(dt)=(F)/(m)=((1)/(2)(t+1))/(4)=(t+1)/(8)`

The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):

`dv=(1)/(8)(t+1)dt\\\int\limits^{v_(2)}_0 \, dv=\int\limits^(2)_(0) {(1)/(8)(t+1)} \, dt`

`v_(2)=(1)/(8)*((t^(2))/(2)+t)` (This is the indefinite integral), the definite one is:

`v_(2)=(1)/(8)*(2+2)=(1)/(2)`