A block of ice with mass 4.00 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force 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?) = ??t?2 + ??t?3 where ??? = 0.200 m/s? and ??? = 0.0200 m/s? (a) Calculate the velocity of the object when ?t? = 4.00 s. (b) Calculate the magnitude of when ?t? = 4.00 s. (c) Calculate the work done by the force during the first 4.00 s of the motion.
Solution 77P Step 1: Data given Condition x(t) = t + t 3 = 0.200 m/s 2 = 0.0200 m/s 3 t = 4.00 s Mass m = 4 kg Step 2 : We need to find the velocity v = d(x)/dt 2 We can write velocity as v = 2 t + 3t Substituting values we can write 2 3 2 v = (2 × 0.200 m/s × 4.00s) + (3 × 0.0200 m/s × (4.00 s) ) v = 1.6 + 0.96 v = 2.56 m/s Hence we have velocity as 2.56 m/s