A 2.0-kg body moving along the x axis has a velocity vx = 5.0 m/s at x = 0. The only force acting on the object is given by Fx = (−4.0x) N, where x is in m. For what value of x will this object first come (momentarily) to rest?
A)4.2 m
B)3.5 m
C)5.3 m
D)6.4 m
E)5.0 m

Question

Quizplus · Accepted Answer

To find the position where the object comes momentarily to rest, we need to find the point where the force acting on it is equal and opposite to the object's initial velocity.

In other words, we need to find the point where:

F(x) = -4.0x = m*a(x) = m*d(v(x))/dt = m*d(v(x))/dx * dx/dt

where m is the mass of the object, a(x) is its acceleration at position x, and v(x) is its velocity at position x.

Since the only force acting on the object is in the x direction and given by F(x) = -4.0x, we can write:

-4.0x = m*d(v(x))/dx * dx/dt

or

-4.0x = m*v(x)*dv(x)/dx

We can separate the variables and integrate both sides:

-4.0x dx = m*v(x) dv(x)

Integrating from x = 0 to x = x_0 (where the object comes momentarily to rest) and from v(0) = 5.0 m/s to v(x_0) = 0, we get:

-2.0x_0^2 = m*v(x_0)^2 - m*v(0)^2

Plugging in the values, we get:

-2.0x_0^2 = 2.0 kg * (0 - 5.0 m/s)^2

Solving for x_0, we get:

x_0 = 3.5 m

A 20-Kg Body Moving Along the X Axis Has a Velocity