A force F=3.0 N+(0.5 N/m)x is used to push a crate from x=0 to x=5.0 m. How much work is done by the force?
The work done by the force is 37.5 J. This can be calculated by using the equation W=F*d, where W is the work done, F is the force, and d is the distance. In this case, W=3.0 N*5.0 m + (0.5 N/m)*(5.0 m)^2 = 37.5 J.
To find the work done by a force, we can use the formula:
Work = Force * Distance * cos(theta)
In this case, the force is given by F = 3.0 N + (0.5 N/m)x. The distance is 5.0 m, and cos(theta) represents the angle between the force vector and the displacement vector.
Since the force is given as a function of position, we need to integrate it over the distance to find the total work. We integrate the force with respect to x:
Work = ∫(F * dx)
= ∫((3.0 N) + (0.5 N/m)x) * dx
Integrating, we get:
Work = (3.0 N)x + (0.5 N/m)(x^2/2) + C
At the starting point x = 0, the work done is zero, as there is no displacement. At the end point x = 5.0 m, we substitute the value into the equation to find the final answer:
Work = (3.0 N * 5.0 m) + (0.5 N/m * (5.0 m)^2/2) + C
= 15.0 Nm + 6.25 Nm + C
Since C represents the constant of integration, we can ignore it as it does not affect the work done. Therefore, the work done by the force is:
Work = 15.0 Nm + 6.25 Nm = 21.25 Nm
To calculate the work done by a force, you need to compute the area under the force-displacement graph. In this case, the force is given by F = 3.0 N + (0.5 N/m)x, where x is the displacement.
Since the force is not constant, you need to integrate the force function over the displacement interval [0, 5.0 m].
The work done (W) is given by the integral:
W = ∫ F dx
W = ∫ (3.0 N + (0.5 N/m)x) dx
W = ∫ (3.0 N) dx + ∫ (0.5 N/m)x dx
W = 3.0 N ∫ dx + 0.5 N/m ∫ x dx
W = 3.0 N (x) + 0.5 N/m (x^2/2) between x = 0 and x = 5.0 m
To calculate the work done, substitute the limits of integration into the equation:
W = 3.0 N (5.0 m) + 0.5 N/m [(5.0 m)^2/2 - (0 m)^2/2]
W = 15.0 N∙m + 0.5 N/m (25 m^2/2)
W = 15.0 N∙m + 6.25 N∙m
W = 21.25 N∙m
Therefore, the work done by the force F is 21.25 N∙m.