How fast is the object moving at x = 1.50 m if it has a speed of 4.00 m/s at x = 1.00 m?

Question

A force with magnitude (5.00 N/m^2)x^2 and directed at a constant angle of 31.0 degrees with the +x axis acts on an object of mass 0.250 kg as the object moves parallel to the x-axis. There are no other forces on the object that have components parallel to the x axis.

How fast is the object moving at x = 1.50 m if it has a speed of 4.00 m/s at x = 1.00 m?

a force of magnitude what??? i dont even understand that part of the question. im used to seeing force in units of Newtons and thats it...that part is incomprehensible to me.

Answer 1

The force has a dependence on the position. The force is (5 N/m^2)x^2. x^2 is the square of position in meters, meaning that the force actually has units of (N/m^2)m^2 = N, which is what you expected.

Now, I'm not sure why the problem specifies that no other forces have components parallel to the x-axis, unless we assume that the object is constrained to move only in the x-direction, so I'll assume that's the case. Then the component of the force in the x-direction is (5x^2)*sin(31) = 2.58x^2. The acceleration due to this force is (2.58x^2) / (0.250) = 10.3x^2, because F = ma ==> a = F/m.

When acceleration is a function of displacement, we integrate (a dx = v dv) to get velocity as a function of displacement. In this case, we integrate (10.3x^2 dx = v dv) which gives us 3.43x^3 = 0.5v^2 + C ==> v = sqrt(6.87x^3) + C. We can solve for C using the given initial conditions. 4 = sqrt(6.87*1^3) + C ==> 4 = sqrt(6.87) + C ==> C = 4 - sqrt(6.87) = 1.38. So when x = 1.5, we have v = sqrt(6.87*1.5^3) + 1.38 = 6.19 m/s.

Answer 2

The last poster is off just a little bit. Most of his work is correct except for the second to last step. The answer should be 7.55 or 7.54 depending on how you round.

Good luck in physics!