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

Answer:

$v_{2} =\frac{1}{2}$

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:

$a=\frac{dv}{dt}$, so:

$F=m*\frac{dv}{dt}\\\frac{dv}{dt}=\frac{F}{m}=\frac{\frac{1}{2}(t+1)}{4}=\frac{t+1}{8}$

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):

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

$v_{2}=\frac{1}{8}*(\frac{t^{2}}{2}+t)$ (This is the indefinite integral), the definite one is:

$v_{2}=\frac{1}{8}*(2+2)=\frac{1}{2}$