Question

A force F-(8.85-4.75t j) N, with t in units of s, is applied to a 2.00 kg object initially at rest. (a) At what time (in s) will the object be moving with a speed of 15.0 m/s? (Round your answer to at least two decimal places.) (b) How far (in m) is the object from its initial position when its speed is 15.0 m/s? (c) What is the total displacement (in m) through which the object traveled at this time? (Express your answer in vector form.)

Answer 1

Here we use simple vector calculation and equations of motion to solve the problem.
Given F = 8.85 i - 4.75 t j mass of object (m) = 2 kg initial velocity = 0 (initially at rost) (a) when velocity v = 15 m/s v = squareroot v^2_x + v^2_y [where v_x = horizontal component of velocity i e i^th component (along x - axis) v_y = jth component of velocity i.e along y axis] now, V = U + at (equation of motion) (where v = final velocity = initial velocity a = acceleration t = time) so, for x - axis: v_x = 0 + a_x t (because u = 0 implies initially at rest) implies v_x = F_x/m t (because F = ma i e force = mass times acceleration) implies v_x = 8.85/2, t = 4.425 t \r\n again, along y - axis: v_y = 0 + a_y t implies v_y = F_y/m t = (- 4.75 t)/2 kg times t = -2.375 t^2 so, v^2 = v^2_x + v^2_y implies 15^2 = (4.425 t)^2

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