Answer to Question #202805 in Mechanics | Relativity for Dell

Question

A block of ice with mass 1.00kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force F to it. A a result, the block moves along the x-axis such that its position as a function of time is given by x(t) = 0.100(m/s2) t2 + 0.0100(m/s3) t3. Calculate the work done by the force during the first 1.00s of the motion. 

Answer 1

m= 1 kg

$x(t) = 0.100 t^2 + 0.0100t^3$

The acceleration of block can be calculated by differentiating position function twice,

$x’(t) = 0.200t + 0.0300t^2 \\ x’’(t) = a = 0.200 + 0.0600t$

Acceleration at t=1 s is:

$a = 0.200 + 0.0600 \times 1 = 0.2600 \;m/s^2$

Force on block can be calculated as:

$F=ma = 1 \;kg \times 0.26 \;m/s^2 = 0.26 \;kg \; m/s^2$

The position of block is at t=1 s.

$x(1) = 0.100 \times 1^2 + 0.0100 \times 1^3 = 0.11 \;m$

Work done by block is:

$W=F \times x \\ = 0.26\times 0.11 = 0.0286 \;J$

Thus, work done is 0.0286 J.