Question

The only force acting on a 2.6 kg body as it moves along the positive x axis has an x component F_x =  −7x N, where x is in meters. The velocity of the body at x = 3.0 m is 8.2m/s.

The only force acting on a 2.6 kg body as it moves along the positive x axis has an x component Fx =-7x N, where x is in meters. The velocity of the body at x = 3.0 m is 8.2 m/s. (a) What is the velocity of the body at x 4.0 m? m/s (b) At what positive value of x will the body have a velocity of 5.0 m/s?

Answer 1

Her, from Newton's second law the expression for the force as follows: F = ma But the acceleration is the rate of change of velocity, then the above equation can be reduced as follows: Fx = ma, d . Q F m ** dx = y dy Integrate the above expression as follows: Ex dx = i v dve 8.2 m/s 3.0mm -12)(L-01 +(1.3)(x+(3.0m)-(13-182 mis

(a) Here, for x =4.0 m the velocity of the body becomes: +(1.3(x+(3.0m)) =(22+63 m/s -(1.35)|(4.0m) –(3.0m)")= (13–168.2.199) * = 48.34 1. - 7 m/s (b) Now, for v = 5.0 m/s the required distance is, -(1.35)(x²–(3.0 m)”)= 15.00 -(1.35)612_(20 m)2)_(5.0 m/s)’ – (8.2 m/s)? (1.35)((x)} -(3.0 m)") = 21.12 (x)²-(3.0 m)² = 15.64 x² = 24.64 x = 4.96 m - 5 m