Question

A boy exerts a force of 11.8 N at 28.0° above the horizontal on a 6.15 kg sled. Find the work done by the boy and the final speed of the sled after it moves 2.75 m, assuming the sled starts with an initial speed of 0.370 m/s and slides horizontally without friction.

Answer 1

Explanation:

It is given that,

Force exerted by a boy, F = 11.8 N

Angle above the horizontal, $\theta=28^{\circ}$

Mass of the sled, m = 6.15 kg

Distance moved, d = 2.75 m

Initial speed, u = 0.37 m/s

Let W is the work done by the boy. Using the expression for the work done to find it as :

$W=Fd\ cos\theta$

$W=11.8\times 2.75\ cos(28)$

W = 28.65 joules

Let v is the final speed of the sled. Using the work energy theorem to find it. It states that the work done is equal to the change in kinetic energy of an object. It is given by :

$W=\dfrac{1}{2}m(v^2-u^2)$

$v=\sqrt{\dfrac{2W}{m}+u^2}$

$v=\sqrt{\dfrac{2\times 28.65}{6.15}+(0.37)^2}$

v = 3.07 m/s

So, the final speed of the sled after it moves 2.75 m is 3.07 m/s. Hence, this is the required solution.