Question

A sled with mass 11.00 kg moves in a straight line on a frictionless horizontal surface. At one point in its path, its speed is 4.00 m/s; after it has traveled a distance 3.00 m beyond this point, its speed is 7.00 m/s. Use the work-energy theorem to find the force acting on the sled, assuming that this force is constant and that it acts in the direction of the sled's motion.

Answer 1

Answer:

The magnitude of the force acting on the sled is 60.5 newtons.

Explanation:

The Work-Energy Theorem states that the work done by the external force applied on the sled ($W$), in joules, is equal to the change of its translational kinetic energy ($\Delta K$), in joules:

$W = \Delta K$ (1)

By definitions of work and translational kinetic energy we expand the equation above:

$F\cdot s = \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})$ (1b)

Where:

$F$ - External force applied on the sled, in newtons.

$s$ - Travelled distance, in meters.

$v_{1}, v_{2}$ - Initial and final velocities, in meters per second.

If we know that $m = 11\,kg$, $v_{1} = 4\,\frac{m}{s}$, $v_{2} = 7\,\frac{m}{s}$ and $s = 3\,m$, then the external force applied on the sled is:

$F = \frac{m\cdot (v_{2}^{2}-v_{1}^{2})}{2\cdot s}$

$F = \frac{(11\,kg)\cdot \left[\left(7\,\frac{m}{s} \right)^{2}-\left(4\,\frac{m}{s} \right)^{2}\right]}{2\cdot (3\,m)}$

$F = 60.5\,N$

The magnitude of the force acting on the sled is 60.5 newtons.