The force on a 0.500-kg particle depends on position such that F(r) = (1.00)x2 i

Question

The force on a 0.500-kg particle depends on position such that F(r) = (1.00)x2 i + ( 4.00)y j N. If the particle starts from rest at r = (0,0), what will be its speed when it reaches the position r = (4, 3) m?
Please show your work fully and clearly. Thanks! The force on a 0.500-kg particle depends on position such that F(r) = (1.00)x2 i + ( 4.00)y j N. If the particle starts from rest at r = (0,0), what will be its speed when it reaches the position r = (4, 3) m?
Please show your work fully and clearly. Thanks! The force on a 0.500-kg particle depends on position such that F(r) = (1.00)x2 i + ( 4.00)y j N. If the particle starts from rest at r = (0,0), what will be its speed when it reaches the position r = (4, 3) m?
Please show your work fully and clearly. Thanks!

Explanation / Answer

F can be resolved into 2 components, Fx along x axis= (1.00)x2 i N and Fy along y axis=  ( 4.00)y j N. Fx moves it along x axis and Fy along y axis. So, work from (0,0) to (4,0) and from (4,0) to (4,3) gives the total amount of work done on the particle . W = Wx + Wy ;where Wx= integral of Fxdx= 1.00*x^3/3 and Wy= integral of Fydy= 4.00*y^2/2

So, W= 1.00/3*4^3 J + 4.00/2*3^2 J
so, W= 39.3 J
Now, 1/2*m*v^2 = 39.3 J { change in kinetic energy is same as the net work done }
so, v^2 = 2/0.500 * 39.3
So, v = 12.5 m/s

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