Question

A 100 kg box as showsn above is being pulled along the x axis by a student. the box slides across a rough surface, and its position x varies with time t according to the equation x=.5t^3 +2t where x is in meters and t is in seconds.
(a) Determine the speed of the box at time t=0
(b) determine the following as functions of time t.

Answer 1

Answer:

a) 2 m/s

b) i) $K.E = 50 (1.5t^2 + 2) ^2$

ii) $F = 3tm$

Explanation:

The function for distance is $x = 0.5t ^3 + 2t$

We know that:

Velocity = $v= \frac{d}{dt} x$

Acceleration = $a= \frac{d}{dt}v$

To find speed at time t = 0, we derivate the distance function:

$x = 0.5 t^3 + 2t\\v= x' = 1.5t^2 + 2$

Substitute t = 0 in velocity function:

$v = 1.5t^2 + 2\\v(0) = 1.5 (0) + 2\\v(0) = 2$

Velocity at t = 0 will be 2 m/s.

To find the function for Kinetic Energy of the box at any time, t.

$Kinetic \ Energy = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times 100 \times (1.5t^2 + 2) ^2\\\\K.E = 50 (1.5t^2 + 2) ^2$

We know that $Force = mass \times acceleration$

$a = v'(t) = 1.5t^2 + 2\\a = 3t$

$F = m \times a\\F= m \times 3t\\F = 3tm$