For the situation described above, what is the work done on the particle by the

Question

For the situation described above, what is the work done on the particle by the force from x = 5.00 m to  x = 22.00 m.

For the situation described above, what is the velocity of the particle when it reaches x = 22.00 m.

For the situation described above, what is the impulse that the particle gives to you as it moves from x = 5.00 m until it reaches x = 22.00 m. Remember that impulse is a vector, so in 1-dimension it can be either positive or negative.

Shown below is a graph of the 1-dimensional, net force as a function of position, F(x), that you exert on a particle of mass 19 kg. Note 1: It may be easier to solve these problems in a different order than they are asked. Note 2: F = 0 at x = 9.190 m. The particle starts at x = 5 m with a velocity of 5.570 m/s and reaches x = 22.00 m under the action of the force in the graph. What is the impulse that you give to the particle as it moves from x = 5.00 m until it reaches x = 22.00 m. Remember that impulse is a vector, so in 1-dimension it can be either positive or negative. For the situation described above, what is the work done on the particle by the force from x = 5.00 m to x = 22.00 m. For the situation described above, what is the velocity of the particle when it reaches x = 22.00 m. For the situation described above, what is the impulse that the particle gives to you as it moves from x = 5.00 m until it reaches x = 22.00 m. Remember that impulse is a vector, so in 1-dimension it can be either positive or negative.

Explanation / Answer

Work done on the particle from x = 5 to x =22 m is given by the area under the Force vs x graph

Area under the graph (from x = 5 to x =22) = 0.5*(9.19-5)*8 - 0.5*(16-9.19)*13 - 13*6 = -105.5 J <----work done

change in K.E = 0.5*m*(Vf^2-Vi^2) = Work done = -105.5 J

where Vi = initial velocty = 5.57 m/s

m = 19 kg

So, 0.5*19*(Vf^2-5.57^2) = -105.5

So, Vf = 4.46 m/s <--------answer

Impulse we give to the particle = m*(Vf-Vi) = 19*(4.46-5.57) = -21.09 kg.m/s <------answer

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