The only force acting on a 2.0 kg body as it moves along a positive x axis has an x component fx = −6 x n, with x in meters. the velocity at x = 3.0 m is 8.0 m/s. (a) what is the velocity of the body at x = 4.0 m? (b) at what positive value of x will the body have a velocity of 5.0 m/s?

a-0

b-2 m

Explanation:

Given mass of the body is 2 kg

Force acting on the body =-6x N

Velocity at x=3m is 8m/s

Now acceleration of the body is -3x

a=-3x

=-3x

vdv=-3xdx

integrating both sides we get

=+c

Now at x=3m v=8m/s

substituting values

c=

Thus

Velocity will be zero after

thus velocity of the body is zero at x=4 m

v=5m/s

25=-3x^2+37

x=2 m

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft