1) The force on a 3.00-kg object as a function of position is shown in the figur

Question

1) The force on a 3.00-kg object as a function of position is shown in the figure.

a) How much work is done by the force as the object moves from x = 0 m to x = 8.0 m?

b) If the object is moving at 2.50 m/s when it is located at x = 0 m, what will its speed be when it reaches x = 8.00 m?

2) A 3 kg particle starts from rest at x = 0 and moves under the influence of a single force F(x) = 5 – 2x + 3x 2 where F(x) is in newtons, x is in meters. Find the work done by the force as the particle moves from x= 0 to x = 4.0 m.

3) A force acts on a 25.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = (3 m/s2 )t 2 + (5 m/s3 )t 3 with x in meters and t in seconds.

a) What is the speed at times t1 = 0 s and t2 = 4.0 s ?

b) Find the work done on the object by the force from t1 = 0 s and t2 = 4.0 s.

Explanation / Answer

1.a) Work Done = Area under the F-x curve = 1*8 + 1/2 *3*3 = 12.5 J

b) Work done = Change in energy

Energy at x(8) - Energy at x(0) = 12.5

Energy at x(8) - 1/2*3*2.5^2 = 12.5

Energy at x(8) = 12.5 + 9.375 = 21.875 J

1/2*3*v^2 = 21.875

v = 3.82 m/s

2) work done = F.dx (from x=0 to x=4) = (5-2x+3x^2) dx = 5x-x^2+x^3

work done  (from x=0 to x=4) = 68 J

3) a)  x = (3 m/s2 )t^2 + (5 m/s3 )t^3

v(t) = dx/dt = 6t + 15t^2

v(0) = 0

v(4) = 264 J

b) work done = change in energy = 1/2*25*264^2 - 0 = 871200 = 871.2 kJ

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