The particle is released at x = 4.5 m with an initial speed of 7.5 m/s, headed i

Question

The particle is released at x = 4.5 m with an initial speed of 7.5 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? (b) What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m? N in the direction. Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 7.5 m/s. (c) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? (d) What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m? N in the direction.

Explanation / Answer

I am assuming that point before 4 is 3 which has not been marked so i take that potential energy is changing from x=3 to x=4m. a) from the graph it is clear that the force experienced by the particle from x=4 to x=3m is given by the slope of the curve as, F= -dU/dx = U2-U1/4-3 = 15 N. As mass of the particle = .82kg so acceleration = F/m = 15/.82 = 18.30 m/s in the region only from x=3 to x=4. Using newtons equation of motion, velocity of the particle at x=3m will be V(x=3)^2= U(x=4)^2 - 2*a*s. As U(x=4) = 7.5m/s, so V(x=3) = sqrt(7.5^2-2*18.3*1) = 4.43283205 m/s. As there is no change in potential energy from x=4 to x=1 m therefore the velocity doesn't change in that region and velocity at x=1m will be 4.4328m/s. b) magnitude and direction of the force as computed in the previous part is equal to 15 N and is directed to the positive x axis. c) we can calculate this using the energy conservation. Energy at starting point x=4.5m = U1 +KE = 20 + .5*.82*(7.5^2) = 43.0625J. Energy at point x=7 is given as U3+ .5*.82*(v^2) = 55 + .5*.82*(v^2) Equating the two we find that velocity^2 of the particle at x=7 is negative which is not possible so the particle cannot reach the point x=7 with the given velocity. To find the turning point, we resort to the method used in the part (a). Force on the particle from x=5 to x=6 m is given by -dU/dx = -slope of the graph F = -[55-20/6-5] =-35N. So acceleration = F/M = -35/.82=-42.68m/s^2 and is directed in the negative x direction. let the distance covered by particle till the velocity of the particle is zero is s so equating the newtons equation of motion. 0^2 = 7.5^2 - 2*42.68*s => s= .653m. Therefore the point where the particle turns is x= 5+.653m = 5.653m. d) as computed in the previous part. The magnitude of the force is 35N and is directed towards the negative x axis. Hope this helps. If you have any further query then please feel free to ask. Wish you award the karma as promised.

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