A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.70kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box.
If the box is initially at rest at x=0, what is its speed after it has traveled 12.0m ?
Take the integral of F(x) which is
18.0x-.5(.530)x^2 then plug in 12.0 for x
=177.84J
so .5mv^2=W since v(initial) is zero so then solve for v
v(final)=((2*177.84)/8.7)^(1/2)
v(final)=6.39m/s
To find the speed of the box after it has traveled 12.0m, we need to determine the work done on the box by the force. The work done can be calculated using the formula:
Work = Change in Kinetic Energy
Since the box starts from rest, its initial kinetic energy is zero. Therefore, the work done will be equal to the final kinetic energy. By applying the work-energy theorem, we can solve for the speed of the box.
1. Determine the work done on the box:
To find the work done, we need to calculate the integral of force with respect to position from x = 0 to x = 12.0m:
Work = ∫ F(x) dx, where F(x) = 18.0N - (0.530N/m)x
Solving the integral, we get:
Work = [18.0x - (0.530N/m)x^2/2] evaluated from x = 0 to x = 12.0
2. Evaluate the work done on the box:
Substituting the values into the equation, we have:
Work = [18.0(12.0) - (0.530N/m)(12.0)^2/2] - [18.0(0) - (0.530N/m)(0)^2/2]
Simplifying further, we have:
Work = 216 - (0.530N/m)(72/2)
Work = 216 - (0.530N/m)(36)
Work = 216 - 19.08N
Therefore, the work done on the box is 216 - 19.08N Joules.
3. Calculate the final kinetic energy:
Since the work done is equal to the final kinetic energy, we have:
Work = Final Kinetic Energy
Substituting the value of work, we get:
216 - 19.08N = (1/2)mv^2, where m = 8.70kg (mass of the box)
4. Solve for the speed of the box:
Rearranging the equation to solve for v (speed), we have:
v^2 = (2(216 - 19.08N))/m
v^2 = (432 - 38.16N)/8.70
Evaluating further, we have:
v^2 = 49.655 - 4.383N
Taking the square root of both sides, we obtain:
v = √(49.655 - 4.383N)
Substituting the value of N = 12.0m, we get:
v = √(49.655 - 4.383(12.0))
Calculating the final answer, we have:
v ≈ 7.44 m/s
Therefore, the speed of the box after traveling 12.0m is approximately 7.44 m/s.
To find the speed of the box after it has traveled 12.0m, we can use the work-energy principle. According to the work-energy principle, the work done on an object is equal to its change in kinetic energy.
The work done by the force acting on the box is given by the integral of the force with respect to displacement:
W = ∫F(x)dx
In this case, the force is given by F(x) = 18.0N - (0.530N/m)x and the displacement is 12.0m. So the work done is:
W = ∫(18.0N - (0.530N/m)x)dx
To find the speed of the box, we need to know the initial kinetic energy of the box, which is zero since it is initially at rest. So the work done on the box is equal to the final kinetic energy of the box:
W = KE_final
Using the work-energy principle, we can calculate the work done on the box:
W = ∫(18.0N - (0.530N/m)x)dx
W = 18.0N(12.0m) - (0.530N/m)(1/2)(12.0m)^2
W = 216N·m - 3.816N·m = 212.184N·m
Since the work done is equal to the final kinetic energy, we can now solve for the speed. The kinetic energy is given by:
KE = (1/2)mv^2
Where m is the mass of the box (8.70kg) and v is the speed of the box. Rearranging the equation, we get:
v^2 = (2KE) / m
Plugging in the values, we have:
v^2 = (2 * 212.184N·m) / 8.70kg
v^2 = 48.76N·m / kg
Finally, we can find the speed of the box by taking the square root of both sides of the equation:
v = √(48.76N·m / kg)
v ≈ 6.98 m/s
Therefore, the speed of the box after traveling 12.0m is approximately 6.98 m/s.