A block of ice with mass 5.50 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force F⃗ to it. As a result, the block moves along the x-axis such that its position as a function of time is given by x(t)=αt2+βt3, where α = 0.210 m/s2 and β = 2.04×10−2 m/s3 . A. Calculate the velocity of the object at time t = 4.50 s .
B. Calculate the magnitude of F⃗ at time t = 4.50 s .
Express your answer to three significant figures.
C. Calculate the work done by the force F⃗ during the first time interval of 4.50 s of the motion.
Express your answer to three significant figures.

A) 3.13 m/s

B) 5.34 N

C) W = 26.9 J

Explanation:

We are told that the position as a function of time is given by;

x(t) = αt² + βt³

Where;

α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³

Thus;

x(t) = 0.21t² + 0.0204t³

A) Velocity is gotten from the derivative of the displacement.

Thus;

v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)

v(t) = 0.42t + 0.0612t²

v(4.5) = 0.42(4.5) + 0.0612(4.5)²

v(4.5) = 3.1293 m/s ≈ 3.13 m/s

B) acceleration is gotten from the derivative of the velocity

a(t) = v'(t) = 0.42 + 2(0.0612t)

a(4.5) = 0.42 + 2(0.0612 × 4.5)

a(4.5) = 0.9708 m/s²

Force = ma = 5.5 × 0.9708

F = 5.3394 N ≈ 5.34 N

C) Since no friction, work done is kinetic energy.

Thus;

W = ½mv²

W = ½ × 5.5 × 3.1293²

W = 26.9 J

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach

the answer is c. the lower legs are levers, and the knees are fulcrums. the ankles hold the loads.