As a 2.0-kg object moves from m to m, the constant resultant force acting on it is equal to N. If the speed of the object at the initial position is 4.0 m/s, what is its kinetic energy at its final position?

Question

As a 2.0-kg object moves from  m to  m, the constant resultant force acting on it is equal to  N. If the speed of the object at the initial position is 4.0 m/s, what is its kinetic energy at its final position?

Quizplus · Accepted Answer

First, we need to find the work done by the constant force on the object using the work-energy principle: 
Work = Change in kinetic energy 
W = (1/2)mv_f^2 - (1/2)mv_i^2 
where m is the mass of the object, v_f is the final velocity, and v_i is the initial velocity.

We know the force acting on the object (F) and the displacement (s) over which it acts, so we can find the work done: 
W = Fs

W = (11 N)(11 m - 8 m) = 33 J

Now we can set up the equation: 
33 J = (1/2)(2 kg)v_f^2 - (1/2)(2 kg)(4.0 m/s)^2

Solving for v_f, we get: 
v_f = 5.13 m/s

Finally, we can find the kinetic energy at the final position: 
KE = (1/2)mv_f^2

KE = (1/2)(2 kg)(5.13 m/s)^2 = 52.9 J ≈ 53 J

Therefore, the answer is B.

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