A 2.7 kg block is initially at rest on a horizontal frictionless surface when a

Question

A 2.7 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by ,

F(x) = (2.8 -x^2)i N


where x is in meters and the initial position of the block is x = 0. (a) What is the kinetic energy of the block as it passes through x = 2.6 m? (b) What is the maximum kinetic energy of the block between x = 0 and x = 2.6 m?

Explanation / Answer

So Newton gives F(x) = m*a(x) so a(x) = (1/m)*(2.8 - x²) The acceleration a(x) = dV(x)/dx (derivative) so: V(x) = ?[a(x)]dx V(x) = ?[(1/m)*(2.8 - x²)]dx = (1/m)* ?[(2.8 - x²)]dx V(x) = (1/m)*[2.8*x - x³/3 + C] C is the arbitrary constant. If x = 0 the velocity is 0 => V(0) = 0 V(0) = (1/m)*[2.8*x - x³/3 + C] = 0 C/m = 0 => C = 0 We have: V(x) = (1/m)*[2.8*x - x³/3] At x = 2 m the velocity is V(2) = 2.44 m/s The kinetic energy is Ek = 0.5*m*V(2)² = 3.585 J The maximum kinetic energy is when the acceleration is 0: This is because the maximum of velocity V(x) = (1/m)*[2.8*x - x³/3] is when the derivative dV(x)/dx = 0 but a(x) = dV(x)/dx so a(x) = 0. a(x) = (1/m)*(2.8 - x²) with 0 a(x) = 0 => (1/m)*(2.8 - x²) = 0 => 2.8 - x² = 0 x = v2.8 = 1.673 m. The velocity is V(1.673) = (1/m)*[2.8*1.673 - (1.673)³/3] = 2.603 m/s so: Emax = 0.5*m*(1.673)² = 4.065 J.

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