Question

Chapter 07, Problem 031 The only force acting on a 1.2 kg body as it moves along the positive x axis has an x component Fx = -4x N, where x is in meters. The velocity of the body at x = 2.0 m is 11 m/s. (a) What is the velocity of the body at x = 4.5 m? (b) At what positive value of x will the body have a velocity of 2.9 m/s? (a) Number Units (b) Number Units

Answer 1

Force acting on body a along x direction is F = -4xN.

Mass of body is M =1.2 kg.

So acceleration of body a from formula f=ma

a=F/M

=-4/1.2

=-3.3m/s²

a) Velocity of particle at distance x=2m is v= 11m/s

So velocity of particle at origin is x=0 is u=?

S distance travelled = 2-0=2m

Now from formula v^{2​​​​​}​​​​- u² = 2as

11²-u² =2(-3.3) 2

121-u² =13.2

u² = 134.2

So velocity at origin u=11.58m/S

Now final Velocity of body at S=4.5m is v=?

Now from formula v² - u² = 2as

v²-(11.58)² =2(-3.3) 4.5

v² -134.2= -29.7

v² =104.5

So velocity at x=4.5m is v=10.22m/S

b) So velocity at origin u=11.58m/S

Now distance at which velocity v=2.9m/s s=?

Now from formula v² - u² = 2as

(2.9)²-(11.58)² =2(-3.3) s

8.41-134.2 = -6. 6(s)

125.79 = 6.6(s)

So the distance at v=2.9m/s is s=19.059m