The graph shows the velocities of two objects of equal mass as a function of tim

Question

The graph shows the velocities of two objects of equal mass as a function of time. Net forces F_A, F_B, and F_C acted on the objects during intervals A, B, and C, respectively. Which one of the following choices is the correct relationship between the magnitudes of the net forces? F_B = FC > FA F_C > F_B > F_A F_A > F_B = F_C F_A = F_B = F_C F_A > F_B > F_C An automobile of mass 1500 kg moving at 20 in s is braked suddenly with a constant braking force of 15000 N. How far does the car travel before stopping? Find the acceleration. Acceleration will be negative. The only force acting horizontally is the breaking force. What is the force needed to pull up a 20.0 kg box upward with an acceleration of 3.0 m/s^2? (It is moving vertically near the surface of the earth).

Explanation / Answer

1.

Using Newton's second law:

Force = mass*acceleration

force is directly proportional to acceleration, since mass is constant for both object.

acceleration = change in velocity/change in time

ab = ac

as you can see change in velocity is same for same amount of change in time.

Here we are only using the fact change.

may be velocity in ac has higher values but change in velocity is equal as the acceleration ab.

So,

Fb = Fc

Now

acceleration in a = 3*change in velocity in Fb/2 *change in time in Fb

acceleration in a = 1.5*acceleration in Fb

So,

Fa > Fb = Fc

Correct option is C.

2.

Force = mass*a

force is negative since it's braking force.

a = -15000/1500 = -10 m/sec^2

u = initial velocoty = 20 m/sec

v = 0 = final velocity

v = u + a*t

t = (v - u)/a

t = -(0 - 20)/10

t = 2 sec

3.

F + ma = mg

F = m (g - a)

F = 20*(9.81 - 3) = 136.2 N

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