A 2-kg box is pushed to the right by a force of 2 N for a distance of 10 m. It h

Question

A 2-kg box is pushed to the right by a force of 2 N for a distance of 10 m. It has an initial velocity of 4 m/s to the right. NOTE: Since this problem gives the DISTANCE traveled, FIRST look at kinetic energy and calculate the NETWORK using wnet F -. After the kinetic energy values are calculated, then calculate the momenta and impulse values. Remember that on quiz and test problems, you will need to decide which values should be calculated first. What is the initial momentum pi of the box? kg m/s Submit Answer Tries o/3 What is the impulse or change in momentum Ap of the box? kg m/s Submit Answer Tries 0/3 What is the final momentum pf of the box? kg m/s Submit Answer Tries 0/3 What is the initial kinetic energy Kj of the box? Submit Answer Tries 0/3 What is the change in kinetic energy AK of the box? Submit Answer Tries o/3 What is the final kinetic energy Kf of the box? Submit Answer Tries 0/3 How long At does it take for the box to travel the distance of 10 m? Submit Answer Tries 0/3

Explanation / Answer

Initial momentum

pi = mvi = 2*4 = 8 kg-m/s

Impulse/change in momentum

Change in KE = W = F*x = 2*10 = 20

0.5*m(vf2 – vi2) = 20

=> vf2 = 20/(0.5*2) + 42

=> vf = 6 m/s

Thus final momentum is:

pf = mvf = 2*6 = 12 kg-m/s

Thus:

change in momentum = 12 – 8 = 4 kg-m/s

Initial KE = ½ * m*vi^2 = ½ * 2 * 42 = 16 J

Final KE = ½ * m*vf^2 = ½ * 2 * 62 = 36 J

Change in KE = 36 – 16 = 20 J

Time taken:

t = impulse/F = 4/2 = 2s

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